Laplace Conversion

Laplace Table

Insert table here

Finding Time Domain Output $y(t)$

1. Transform $x(t)$ and $h(t)$ into Laplace domain
2. Find product $Y(s)=X(s)H(s)$
3. Take inverse Laplace transform $Y(s)$

Input as Delta Function $\delta (t)$

$$x(t)=\delta (t)$$ Then $X(s)=1$, so $Y(s)=H(s)$.

Input as Step Function $u(t)$

$$x(t)=u(t)$$ Then $X(s)=\frac{1}{s}$, so $Y(s)=\frac{H(s)}{s}$.

LTI System Properties

LTI = Linear Time Invariant.

• LTI systems are linear. Given system $F\{\}$ and signals $x_1(t)$, $x_2(t)$ etc
  • LIT is Additive: $$F\{x_1(t)+x_2(t)\}=F\{x_1(t)\}+F\{x_2(t)\}$$
  • LTI is scalable (or homogeneous) $$F\{\alpha x_1(t)\}=\alpha F\{x_1(t)\}$$
• LTI is time-invariant, ie, if output $y(t)=F\{x_1(t)\}$ then:
  • $$y(t-\tau )=F\{x_1(t-\tau )\}$$

3 - Poles and Zeros

General Transfer Function as 2 polynomials

$$H(s)=\frac{b_0{s}^M+b_1{s}^{M-1}+\cdots +b_{M-1}s+b_M}{a_0{s}^N+a_1{s}^{N-1}+\cdots +a_{N-1}s+a_N}$$

Factorised Transfer Function

$$H(s)=K\frac{(s-z_1)(s-z_2)\cdots (s-z_M)}{(s-p_1)(s-p_2)\cdots (s-p_N)}$$ Is factorised and rewrite as a ratio of products: $$=K\frac{\prod {}_{t=1}^{M}s-z_t}{\prod {}_{t=1}^{N}s-p_t}$$

Real system as real

$$M\le N$$ Where the numerator i a $M$th order polynomial with coefficients $b$s and the denominator is a $N$th order polynomial with coefficients $a$s. For a system to be real, the order of the numerator polynomial must be no greater than the order of the denominator polynomial, ie: $M\le N$.

Zero Definition

Roots z of the numerator. When $s=$ any $z$, $H(s)=0$

Pole Definition

Poles p of the denominator. When $s=$ any $p$, $H(s)$ approaches $\inf$

Transfer Function Gain

K is the overall transfer function gain. (Coefficient of $s^M$ and $s^N$ is 1.)

Stable System

A system is considered stable if its impulse response tends to zero or a finite value in the time domain.

Requires all real components to be negative (on the left hand side of the complex s-plane of a pole-zero plot (left if the imaginary s axis)).

Components to Response

Real Components $\Rightarrow$ Exponential Response $|$ Imaginary $\Rightarrow$ angular frequency of oscillating responses.

4 - Analog Frequency Response

Frequency Response

Frequency response of a system = output in response to sinusoid input of unit magnitude and specified frequency, $\omega$. Response is measured as magnitude and phase angle.

Continuous Fourier Transform

$$F(jw)=\int {}_{t=0}^{\infty }f(t)e^{-j\omega t}dt$$

Laplace transform evaluated on the imaginary s-axis at some frequency $s=j\omega$.

$\omega =$ radial frequency, $\frac{rad}{s}$

Inverse Fourier Transform

$$f(t)=\frac{1}{2\pi }\int {}_{\omega =-\infty }^{\infty }F(j\omega )e^{j\omega t}d\omega$$

Magnitude of Frequency Response (MFR) $|H(j\omega )|$

$$\left|H(j\omega )\right|=\left|K\right|\frac{{\prod }_{i=1}^M\left|j\omega -z_i\right|}{{\prod }_{i=1}^N\left|j\omega -p_i\right|}$$

In words, the magnitude of the frequency response (MFR) $|H(j\omega )|$ is equal to the gain multiplied by the magnitudes of the vectors corresponding to the zeros, divided by the magnitudes of the vectors corresponding to the poles.

Phase Angle of Frequency Response (PAFR) $\angle H(j\omega )$ - $K>0$

$$\angle H(j\omega )=\sum _{i=1}^M\angle (j\omega -z_i)-\sum _{i=1}^N\angle (j\omega -p_i)$$

Phase Angle of Frequency Response (PAFR) $\angle H(j\omega )$ - $K<0$

$$\angle H(j\omega )=\sum _{i=1}^M\angle (j\omega -z_i)-\sum _{i=1}^N\angle (j\omega -p_i)+\pi$$

In words, the phase angle of the frequency response (PAFR) $\angle H(j\omega )$ is equal to the sum of the phases of the vectors corresponding to the zeros, minus the sum of the phases of the vectors correspond to the poles, plus $\pi$ if the gain is negative.

Each phase vector is measured from the positive real s-axis (or a line parallel to the real s-axis if the pole or zero is not on the real s-axis).

5 - Analog Filter Design

Ideal Filters

Each ideal filter has unambiguous pass bands, which are ranges of frequencies that pass through the system without distortion, and stop bands, which are ranges of frequencies that are rejected and do not pass through the system without significant loss of signal strength. The transition band between stop and pass bands in ideal filters has a size of 0; transitions occur at single frequencies.

Realisability

System starts to respond to input before input is applied. Non-zero for $t<0$.

Causality

Output depends only on past and current inputs, not future inputs.

Realising Filters

Realise as we seek smooth behaviour.

• Drop $h_i(t)$ for $t<0$ ($h_i(t)u(t)$)
  • Would not get suitable behaviour in frequency domain, as discarded 50% of system energy
• But can tolerate delays
  • So shift sinc to the right
  • Time domain shift = scaling by complex exponential in laplace
  • True in fourier transform, so delay in time maintains magnitude but changes phase of frequency response
• Truncate
  • As can't wait for infinity, so truncate impulse response.

Gain $G_{dB}$ (linear $\to$ dB)

$$G_{dB}=20lo{g}_{10}(G_{linear})$$

Gain $G_{linear}$ (dB $\to$ linear)

$$G_{linear}=10^{\frac{G_{dB}}{20}}$$

Transfer Function of Nth Order Butterworth Low Pass Filter

$$H(s)=\frac{{\omega }_c^N}{{\prod }_{n=1}^N(s-p_n)}$$

Butterworth = Maximally flat in pass band (freq response magnitudes are flat as possible for given order)

• $p_n$ = nth pole
  • = $j\omega {}_c{e}^{\frac{j\pi }{2N}(2n-1)}$
  • = $-\omega {}_{c}sin(\frac{\pi (2n-1)}{2N})+j{\omega }_{c}cos(\frac{\pi (2n-1)}{2N})$
  • Form semi-circle to left of imaginary s-axis
• $\omega {}_c$ = half-power cut-off frequency
  • Frequency where filter gain is $G_{linear}=\frac{1}{\sqrt{2}}$ or $G_{dB}=-3dB$

Frequency Response of common Low pass Butterworth filter

$$H(j\omega )=\frac{1}{\sqrt{1+(\frac{w}{w_c}{)}^{2N}}}$$

Increasing order improves approximation of ideal behaviour

Normalised Frequency Response of common Low pass Butterworth filter

$$H(j\omega )=\frac{1}{\sqrt{1+w^{2N}}}$$

To convert normalised frequency form to non-normalised = multiply $\omega$ by the actual ${\omega }_c$

Minimum Order for Low Pass Butterworth

$$N=\lceil \frac{log(\frac{10^{-\frac{G_s}{10}}-1}{10^{-\frac{G_p}{10}}-1})}{2log(\frac{{\omega }_s}{{\omega }_p})}\rceil$$

Round up as want to over-satisfy not under-satisfy

Low pass Butterworth Cut-off frequency ${\omega }_c$ (Pass)

$${\omega }_c=\frac{{\omega }_p}{(10^{-\frac{G_p}{10}}-1{)}^{\frac{1}{2N}}}$$

Gain in dB

Low pass Butterworth Cut-off frequency ${\omega }_c$ (Stop)

$${\omega }_c=\frac{{\omega }_s}{(10^{-\frac{G_s}{10}}-1{)}^{\frac{1}{2N}}}$$

Gain in dB

6 - Periodic Analogue Functions

Exponential Representation from Trigonometric representation

$$e^{jx}=\cos x+j\sin x$$

Trigonometric from exponential - Real (cos)

$$\cos x=Re\{e^{jx}\}=\frac{e^{jx}+e^{-jx}}{2}$$

Trigonometric from exponential - Imaginary (cos)

$$\sin x=Im\{e^{jx}\}=\frac{e^{jx}+e^{-jx}}{2j}$$

Fourier Series

$$x(t)=\sum _{k=-\infty }^{\infty }{X}_k{e}^{jk{\omega }_{0}t}$$ Period signal = sum of complex exponentials.

Fundamental frequency $f_0$, such that all frequencies in signal are multiples of $f_0$.

Fundamental period $T_0=1/f_0$

$w_0=2\pi f_0=2\pi /T_0$

Fourier spectra only exist at harmonic frequencies (ie integer multiples of fundamental frequency)

Fourier Coefficients

$$X_k=\frac{1}{T_0}{\int }_{T_0}x(t)e^{-jk{\omega }_{0}t}dt$$

Important property of Fourier series is how is represents real signals $x(t)$.

• Even magnitude spectrum $\to |X_k|=|X_{-k}|$
• Odd phase spectrum = $\to \angle X_k=-\angle X_{-k}$

Fourier Series of Periodic Square Wave (Example)

$$x(t)=\sum _{k=-\infty }^{\infty }\frac{A\tau }{T_0}sinc(k{\omega }_0\frac{\tau }{2})e^{jk{\omega }_{0}t}$$

Where $X_k=\frac{A\tau }{T_0}sinc(k{\omega }_0\frac{\tau }{2})$

Output of LTI system from Signal with multiple frequency components

$$y(t)=\sum _{k=-\infty }^{\infty }H(jk{\omega }_0)X_k{e}^{jk{\omega }_{0}t}$$

Or in other words:

$$Y_k=H(jk{\omega }_0)X_k$$

The output of an LTI system due to a signal with multiple frequency components can be found by superposition of the outputs due to the individual frequency components. IE system will change amplitude and phase of each frequency in the input.

Filtering Periodic Signal (Example 6.2)

See example 6.2 below...

7 - Computing with Analogue Signals

This topic isn't examined as it is MATLAB

8 - Signal Conversion between Analog and Digital

Digital Signal Processing Workflow

See diagram:

• Low pass filter applied to time-domain input signal $x(t)$ to limit frequencies
• An analogue-to-digital converter (ADC) samples and quantises the continuous time analogue signal to convert it to discrete time digital signal $x[n]$.
• Digital signal processing (DSP) performs operations required and generates output signal $y[n]$.
• A digital-to-analogue converter (DAC) uses hold operations to reconstruct an analogue signal from $y[n]$
• An output low pass filter removes high frequency components introduced by the DAC operation to give the final output $y(t)$.

Sampling

Convert signal from continuous-time to discrete-time. Record amplitude of the analogue signal at specified times. Usually sampling period is fixed.

Oversample

Sample too often, use more complexity, wasting energy

Undersample

Not sampling often enough, get aliasing of our signal (multiple signals of different frequencies yield the same data when sampled.)

Aliasing

Multiple signals of different frequencies yield the same data when sampled.

If we sample the black sinusoid at the times indicated with the blue marker, it could be mistaken for the red dashed sinusoid. This happens when under-sampling, and the lower signal is called the alias. The alias makes it impossible to recover the original data.

Nyquist Rate

$${\omega }_s=2{\omega }_B$$

Minimum ant-aliasing sampling Frequency.

Frequencies above this ${\omega }_s\ge 2{\omega }_B$ remain distinguishable.

Quantisation

The mapping of continuous amplitude levels to a binary representation.

IE: $W$ bits then there are $2^W$ quantisation levels. ADC Word length $=W$.

Continuous amplitude levels are approximated to the nearest level (rounding). Resulting error between nearest level and actual level = quantisation noise

Data Interpolation

Convert digital signal back to analogue domain, reconstruct continous signal from discrete time series of points.

Hold Circuit

Simplest interpolation in a DAC, where amplitude of continuous-time signal matches that of the previous discrete time signal.

IE: Hold amplitude until the next discrete time value. Produces staircase like output.

Resolution

$$\frac{1}{2^W}\times 100\%$$

Space between levels, often represented as a percentage.

For $W$-bit DAC, with uniform levels

Dynamic range

$$20lo{g}_{10}{2}^W\approx 6WdB$$

Range of signal amplitudes that a DAC can resolve between its smallest and largest (undistorted) values.

9 - Z-Transforms and LSI Systems

LSI Rules

Linear Shift-Invariant

Common Components of LSI Systems

For digital systems, only need 3 types of LSI circuit components.

1. A multiplier scales the current input by a constant, i.e., $y[n]=b[1]x[n]$.
2. An adder outputs the sum of two or more inputs, e.g., $y[n]=x_1[n]+x_2[n]$.
3. A unit delay imposes a delay of one sample on the input, i.e, $y[n]=x[n-1]$.

Discrete Time Impulse Function

Impulse response is very similar in digital domain, as it is the system output when the input is an impulse.

Impulse Response Sequence

$$h[n]=\mathcal{F}\{\delta [n]\}$$

LSI Output

$$y[n]=\sum _{k=-\infty }^{\infty }x[k]h[n-k]=x[n]\ast h[n]=h[n]\ast x[n]$$

Discrete Convolution of input signal with the impulse response.

Z-Transform

$$\mathcal{Z}\{f[n]\}=F(z)=\sum _{k=0}^{\infty }f[k]z^{-k}$$

Converts discrete-time domain function $f[n]$ into complex domain function $F(z)$, in the z-domain Assume $f[n]$ is causal, ie $f[n]=0,\forall n<0$

Discrete time equivalent to Laplace Transform. However can be written by direct inspection (as have summation instead of intergral). Inverse equally as simple.

Z-Transform Examples

Simple examples...

Binomial Theorem for Inverse Z-Transform

$$\sum _{n=0}^{\infty }{a}^n=\frac{1}{1-a}$$

Cannot always find inverse Z-tranform by immediate inspection, in particular if the Z-transform is written as a ratio of polynomials of z. Can use Binomial theorem to convert into single (sometimes infinite length) polynomial of $z$

Z-Transform Properties

Linearity, Time Shifting and Convolution

Sample Pairs

See example

Z-Transform of Output Signal

$$Y(z)=\mathcal{Z}\{y[n]\}=\mathcal{Z}\{x[n]\ast h[n]\}=\mathcal{Z}\{x[n]\}\mathcal{Z}\{h[n]\}=X(z)H(z)\Rightarrow Y(z)=X(z)H(z)$$

Where $H(z)$ = Pulse Transfer Function (as it is also the system output when the time-domain input is a unit impulse.) but by convention can refer to $H(z)$ as the Transfer Function

Finding time-domain output $y[n]$ of an LSI System

Transform, product, inverse.

1. Transform $x[n]$ and $h[n]$ into z-domain
2. Find product $Y(z)=X(z)H(z)$
3. Taking the inverse Z-transform of $Y(z)$

Difference Equation

Time domain output $y[n]$ directly as a function of time-domain input $x[n]$ as well as previous time-domain outputs $x[n-k]$ (ie can be feedback).

Z-Transform Table

See table...

10 - Stability of Digital Systems

Z-Domain Transfer Function

$$H(z)=\frac{b[M]z^{-M}+b[M-1]z^{1-M}+\cdots +b[1]z^{-1}+b[0]}{a[N]z^{-N}+a[N-1]z^{1-N}+\cdots +a[1]z^{-1}+1}$$

Negative powers of z.

No constraint on $M$ and $N$to be real (unlike analogue) but often assume $M=N$

General Difference Equation

$$y[n]=\sum _{k=0}^{M}b[k]x[n-k]-\sum _{k=0}^{N}a[k]y[n-k]$$

Poles and Zeros of Transfer Function

$$H(z)=K\frac{(z-z_1)(z-z_2)\cdots (z-z_M)}{(z-p_1)(z-p_2)\cdots (z-p_M)}=K\frac{{\prod }_{i=1}^{M}z-z_i}{{\prod }_{i=1}^{N}z-p_i}$$

• Coefficient of each $z$ in this form is 1.
• Poles $p_i$ and zeros $z_i$ carry same meaning as analogue
• Unfortunately symbol for variable $z$ and zeros $z_i$ are very similar (take care)
• Insightful to plot

Bounded Input and Bounded Output (BIBO) Stability

Stable if bounded input sequence yields bounded output sequence.

A system is BIBO stable if all of the poles lie inside the $|z|=1$ unit circle

A system is Conditionally stable if there is atleast 1 pole directly on the unit circle.

Explanation:

• An input sequence $x[n]$ is bounded if each element in the sequence is smaller than some value $A$.
• An output sequence $y[n]$ corresponding to $x[n]$ is bounded if each element in the sequence is smaller than some value $B$.

11 - Digital Frequency Response

LSI Frequency Response

Output in response to a sinusoid input of unit magnitude and some specified frequency. Shown in two plots (magnitude and phase) as a function of input frequency.

Discrete-Time Fourier Transform (DTFT) - Digital Frequency Response

$$F(e^{j\Omega })=F(\Omega )=\sum _{k=0}^{\infty }f[k]e^{-jk\Omega }$$

Where angle $\Omega$ is the angle of th unit vector measured from the positive real $z$-axis. Denotes digital radial frequency, measured in radians per sample $\frac{rad}{sample}$

$F(e^{j\Omega })$ as spectrum of $f[n]$ (frequency response).

Convention of writing DTFT includes $F(e^{j\Omega })$ or simply $F(\Omega )$

Derivation: Using Z-Transform Definition. $$\mathcal{Z}\{f[n]\}=F(z)=\sum _{k=0}^{\infty }f[k]z^{-k}$$

Let $z$ be polar coords ($z=r{e}^{j\Omega }$), ie magnitude to r, angle $\Omega$. Hence rewrite $$F(r{e}^{j\Omega })=\sum _{k=0}^{\infty }f[k]r^{-k}{e}^{-jk\Omega }$$

Then let $r=1$, so that any point lies on the unit circle. $$F(e^{j\Omega })=F(\Omega )=\sum _{k=0}^{\infty }f[k]e^{-jk\Omega }$$

Inverse Discrete-Time Fourier Transform (Inverse DTFT)

$$f[k]=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }F(e^{j\Omega })e^{jk\Omega }d\Omega$$

LSI Transfer Function

$$H(e^{j\Omega })=K\frac{{\prod }_{i=1}^M{e}^{j\Omega }-z_i}{{\prod }_{i=1}^N{e}^{j\Omega }-p_i}$$

$H(e^{j\Omega })$ is a function of vectors from the system's poles and zeros to the unit circle at angle $\Omega$.Thus from pole-zero plot, can geometrically determine magnitude and phase of frequency response.

Magnitude of Frequency Response (MFR) $|H(e^{j\Omega })|$

$$\left|H(e^{j\Omega })\right|=\left|K\right|\frac{{\prod }_{i=1}^M\left|e^{j\Omega }-z_i\right|}{{\prod }_{i=1}^N\left|e^{j\Omega }-p_i\right|}$$

In words, the magnitude of the frequency response (MFR) $|H(e^{j\Omega })|$ is equal to the gain multiplied by the magnitudes of the vectors corresponding to the zeros, divided by the magnitudes of the vectors corresponding to the poles.

Repeats every $2\pi$ as Eulers Formula. Due to symettery of poles and zeros about real $z$-axis, frequency response is symmetric about $\Omega =\pi$, so only need to find over one interval of $\pi$

Phase Angle of Frequency Response (PAFR) $\angle H(e^{j\Omega })$ - $K>0$

$$\angle H(e^{j\Omega })=\sum _{i=1}^M\angle (e^{j\Omega }-z_i)-\sum _{i=1}^N\angle (e^{j\Omega }-p_i)$$

In words, the phase angle of the frequency response (PAFR) $\angle H(e^{j\Omega })$ is equal to the sum of the phases of the vectors corresponding to the zeros, minus the sum of the phases of the vectors correspond to the poles, plus $\pi$ if the gain is negative.

Repeats every $2\pi$ as Eulers Formula. Due to symettery of poles and zeros about real $z$-axis, frequency response is symmetric about $\Omega =\pi$, so only need to find over one interval of $\pi$

Example 11.1 - Simple Digital High Pass Filter

See image...

12 - Filter Difference equations and Impulse responses

Z-Domain Transfer Function

$$H(z)=\frac{b[M]z^{-M}+b[M-1]z^{1-M}+\cdots +b[1]z^{-1}+b[0]}{a[N]z^{-N}+a[N-1]z^{1-N}+\cdots +a[1]z^{-1}+1}$$

General Difference Equation

$$y[n]=\sum _{k=0}^{M}b[k]x[n-k]-\sum _{k=0}^{N}a[k]y[n-k]$$

Real coefficients $b[\dot{]}$ and $a[\dot{]}$ are the same. (Note $a[0]$ = 1, so no coefficient corresponding to $y[n]$).

Easier to convert directly between transfer function $H(z)$ (with negative powers of z) and the difference equation for output $y[n]$, ideal for implementation of the system. (rather than use time-domain impulse response $h[n]$)

Example 12.1 Proof y[n] can be obtained directly from H[z]

See image...

Order of a filter

$$order=max(N,M)$$

Taps in a filter

Minimum number of unit delay blocks required. Equal to the order of the filter.

Example 12.2 Filter Order and Taps

See example...

Tabular Method for Difference Equations

Given a difference equation, and its input x[n], can write specific output y[n] using tabular method.

1. Starting with input $x[n]$, make a column for every input and output that appears in difference equation
2. ASsume every output and delayed input is initially zero (ie the filter is causal, initially no memory, hence system is quiescent)
3. Fill in column for $x[n]$ with given system input for all rows needed, and fill in delayed versions of $x[n]$
4. Evaluate $y[0]$ from inital input, and propagate the value of y[0] to delayed outputs (as relavent)
5. Evaluate $y[1]$ from $x[\dot{]}$s and $y[0]$
6. Continue evaluating output and propagating delayed outputs.

Can be alternative method for finding time-domain impulse response $h[n]$

Example 12.3 Tabular Method Example

See example

Infinite Impulse Response (IIR) Filters

IIR filters have infinite length impulse responses because they are recursive (ie feedback terms associated with non-zero poles in transfer function, hence $y[n-k]$ exists.)

Standard transfer function and difference equation can be used to represent. $$H(z)=\frac{b[M]z^{-M}+b[M-1]z^{1-M}+\cdots +b[1]z^{-1}+b[0]}{a[N]z^{-N}+a[N-1]z^{1-N}+\cdots +a[1]z^{-1}+1}$$ $$y[n]=\sum _{k=0}^{M}b[k]x[n-k]-\sum _{k=0}^{N}a[k]y[n-k]$$

Not possible to have a linear phase response (so there are different delays associated with different frequencies, and they are not always stable (depending on the exact locations of poles.))

IIR filters are more efficient than FIR designs at controlling gain of response.

Although response is technically infinite, in practice decays towards zero or can be truncated to zero (assume response is $h[n]=0$ beyond some value $n$)

Example 12.4 IIR Filter

See example

Finite Impulse Response (FIR) Filters

FIR Filter are none recursive (ie, no feedback components), so a[k] = 0 for k!=0.

Finite in length, and strictly zero beyond that ($h[n]=0$ for $n>M$). Therefore the number of filter taps dictates the length of an FIR impulse response

Since there is no feedback, can write impulse response $h[n]$ as: $$h[n]=b[n]$$

FIR Difference Equation

$$y[n]=\sum _{k=0}^{M}b[k]x[n-k]$$

FIR Transfer function

$$H(z)=b[M]z^{-M}+b[M-1]z^{1-M}+\cdots +b[1]z^{-1}+b[0]$$

Simplified from general differernce equation tranfer function.

FIR Transfer Function - Roots

$$H(z)=\frac{b[M]+b[M-1]z+\cdots +b[1]z^{M-1}+b[0]z^M}{z^M}=K\frac{{\prod }_{k=0}^M(z-z_k)}{z^M}$$

More convenient to work with positive powers of z, so multiply top and bottom by $z^M$ then factor.

FIR Stability

FIR FILTERS ARE ALWAYS STABLE. As in transfer function, all M poles are all on the origin (z =0) and so always in the unit circle.

FIR Linear Phase Response

Often have a linear phase response. The phase shift at the output corresponds to a time delay.

FIR Filter Example

See example 12.5

Ideal Digital Filters

Four main types of filter magnitude responses (defined over $0\le \Omega \le \pi$, mirrored over $\pi \le \Omega <2\pi$ and repeated every $2\pi$)

• Low Pass - pass frequencies less than cut-off frequency ${\Omega }_c$ and reject frequencies greater.
• High Pass - rejects frequencies less than cut-off frequency ${\Omega }_c$ and pass frequencies greater.
• Band Pass - Passes frequency within specified range, ie between ${\Omega }_1$ and ${\Omega }_2$, and reject frequencies that are either below or above the band within $[0,\pi ]$
• Band Stop - Rejects frequency within specified range, ie between ${\Omega }_1$ and ${\Omega }_2$, and passes all other frequencies within $[0,\pi ]$

Ideal response appear to be fundamentally different from ideal analogue, however we only care over fundamental band $[-\pi ,\pi )$ where behaviour is identical

Realising Ideal Digital Filters

Use poles and zeros to create simple filters. Only need to consider response over the fixed $[-\pi ,\pi )$ band.

Key Concepts:

• To be physically realisable, complex poles and zeros need to be in conjugate pairs

• Can place zeros anywhere, so will often place directly on unit circle when frequency / range of frequency needs to be attenuated

• Poles on the unit circle should generally be avoided (conditionally stable). Can try to keep all poles at origin so can be FIR, otherwise IIR, so feedback. Poles used to amplify response in the neighbourhood of some frequency.

• Low Pass - zeros at or near $\Omega =\pi$, poles near $\Omega =0$ which can amplify maximum gain, or be used at a higher frequency to increase size of pass band.

• High Pass - literally inverse of low pass. Zeros at or near $\Omega =0$, poles near $\Omega =\pi$ which can amplify maximum gain, or be used at a lower frequency to increase size of pass band.

• Band Pass - Place zeros at or near both $\Omega =0$ and $\Omega =\pi$; so must be atleast second order. Place pole if needed to amplify the signal in the neighbourood of the pass band.

• Band Stop - Place zeros at or near the stop band. Zeros must be complex so such a filter must be atelast second order. Place poles at or near both $\Omega =0$ and $\Omega =\pi$ if needed.

Example 12.6 - Simple High Pass Filter Design

See diagram

13 - FIR Digital Filter Design

Discrete Time Radial Frequency

$$\Omega =\frac{\omega }{f_s}=\frac{2\pi f}{f_s}$$

As long as $\Omega \le \pi$ - otherwise there will be an alias at a lower frequency. So $2f\le f_s$ to avoid aliasing.

Realising Ideal Digital Filter

Aim is to get as close as possible to ideal behaviour. But when using Inverse DTFT, the ideal impulse response is $\frac{{\Omega }_c}{\pi }sinc(n{\Omega }_c)$.

This is analogous to the inverse Fourier transform of the ideal analogue low pass frequency response.

Sampled a scaled sinc function, non-zero values for $n<0$. So needs to respond to an input before the input is applied, thus unrealisable.

Practical Digital Filters

Good digital low pass filter will try to realise the (unrealisable) ideal response. Will try to do this with FIR filters (always stable, tend to have greater flexibility to implement different frequency responses).

• Need to induce a delay to capture most of the ideal signal energy in causal time, ie: use $h_i[n-k]u[n]$
• Truncate response to delay tolerance $k_d$, such that $h[n]=0$ for $n>k_d$. Also limits complexity of filter: shorter = smaller order
• Window response, scales each sample, attempt to mitigate negative effects of truncation

Windowing

Window Method - design process: start with ideal $h_i[n]$ and windowing infinite time-domain response to obtain a realsiable $h[n]$ that can be implemented.

Windowing Criteria

• Main Lobe Width - Width in frequency of the main lobe.
• Roll-off rate - how sharply main lobe decreases, measured in dB/dec (db per decade).
• Peak side lobe level - Peak magnitude of the largest side lobe relative to the main lobe, measured in dB.
• Pass Band ripple - The amount the gain over the pass band can vary about unity $1-{\delta }_p/2$ and $1+{\delta }_p/2$
• Pass Band Ripple Parameter, dB- $r_p=20lo{g}_{10}(\frac{1+{\delta }_p/2}{1-{\delta }_p/2})$
• Stop Band ripple - Gain over the stop band, must be less than the stop band ripple ${\delta }_s$
• Transition Band - ${\Omega }_{\Delta }={\Omega }_s-{\Omega }_p$

Practical FIR Filter Design Example 13.2

See example...

Specification for FIR Filters Example 13.3

See example...

14 - Discrete Fourier Transform and FFT

Discrete Fourier Transform DFT

$$X[k]=\sum _{n=0}^{N-1}x[n]e^{-jnk\frac{2\pi }{N}}$$

For $k=\{0,1,\mathrm{2...},N-1\}$

This is Forward discrete Fourier transform. (Not discrete time transform, but samples of it over interval $[0,2\pi )$

Explanation:

Discrete-time Fourier Transfomr (DTFT), takes discrete time signal, provides continous spectrem that repeats every $2\pi$. Defined for infinite length sequency $x[n]$, gives continous spectrum with values at all frequencies.

$$X(e^{j\Omega })=\sum _{n=0}^{\infty }x[n]e^{jn\Omega }$$

Digital often has finite length sequences. (Also inverse DTFT, uses intergration thus approximated). So assume sequence $x[n]$ is length $N$.

$$X(e^{j\Omega })=\sum _{n=0}^{N-1}x[n]e^{jn\Omega }$$

Sample spectrum $X[e^{j\Omega }]$. Repeats every $2\pi$, can sample over $[0,2\pi )$.

Take same number of samples in frequency domain as length of time domain signal.So $N$ evenly spaced samples of $X(e^{j\Omega })$. (Aka bins)

Occur at fundemental frequency ${\Omega }_0=2\pi /N$ $$\Omega =\left\{0,\frac{2\pi }{N},\frac{4\pi }{N},\cdots ,\frac{2\pi (N-1)}{N}\right\}$$

Substitude into the DTFT.

$$X[k]=\sum _{n=0}^{N-1}x[n]e^{-jnk\frac{2\pi }{N}}$$

For $k=\{0,1,\mathrm{2...},N-1\}$

$$f_k=k\frac{f_s}{N},0\le k\le N-1$$

Inverse DFT

$$x[n]=\frac{1}{N}\sum _{k=0}^{N-1}X[k]e^{jnk\frac{2\pi }{N}},n=\left\{0,1,2,\cdots ,N-1\right\}$$

Example 14.1 DFT of Sinusoid

See example

Zero Padding

Artificially increase the length of the time domain signal $x[n]$ by adding zeros to the end to see more detail in the DTFT as DFT provides sampled view of DTFT, only see DTFT at $N$ frequencies.

Example 14.2 Effect of Zero Padding

See example

Fast Fourier Transform FFT

Family of alogrithms that evaluate DFT with complexity of $O(Nlo{g}_{2}N)$ compared to $O(N^2)$. Achieved with no approximations.

Details are beyond module, but can be used in matlab with fft function.