Design digital filter:
1. Draw unit circle+ mark freqs
𝑓
𝑗2𝜋𝑓
𝑠
2. Place poles + zeros (using α𝑒
𝑓
where α usually < 1, 2𝜋 = angle)
General form of transfer func.
Linear phase if zeros on
Symmetry about x-axis – if
H(z)=N(z)/D(z)
unit circle and/or
pole exists at z=Aejα there must
Unit circle
Zeros
when
H(z)0
Shift theorem: zeros/poles at origin
be one at z=Ae-jα
Poles when H(z)∞
𝑌(𝑧) = 𝑧 −𝑚 𝑋(𝑧)
2𝜋𝑝
Angle of pth
When zn=1, 𝑧 = 𝑒 𝑗 𝑛 where
𝑝
Frequency response of a
shift of m samples
root = 2𝜋
𝑓
𝑛
from X(z) to Y(z)
p starts from 0, 1..., n
digital filter = H(ej2πfT)
Z-transform:
𝑗2𝜋
∞
𝑧 = 𝑒 𝑓𝑠
Fourier transform =
e.g. 0.7y[n-1]0.7z-1Y(z)
𝑋(𝑧) = ∑ 𝑥[𝑛]𝑧 −𝑛
X(ej2πfT) of X(z)
Convolution theorem:
∞
𝑛=−∞
Discrete convolution:
Latches delay
∞
𝑣𝑜 (𝑡) = ∫ ℎ(𝑡 − 𝜏)𝑣𝑖 (𝜏)𝑑𝜏
𝑉𝑜 (𝑗𝜔) = 𝐻(𝑗𝜔)𝑉𝑖 (𝑗𝜔)
signal by T
(j2πf for cyclical frequency response)
𝑣𝑜 [𝑛] = ∑ ℎ[𝑛 − 𝑝]𝑣𝑖 [𝑝]
−∞
𝑝=−∞
If p1 is between 0 and ½N, plot in DFT;
Amplitude of
4. Spectrum frequency represented if not, use periodicity property of DFT:
𝐴𝑁
𝑝𝑓
peak =
If a peak occurs at p, it will occur at p±N,
by pth harmonic, 𝑓𝑝 = 𝑠
2
𝑁
p±2N... do same for all p1, p2...
𝑁−1
−𝑗2𝜋𝑛𝑝⁄
1. Sample original signal (with duration T seconds)
𝑁
𝑋[𝑝] = ∑ 𝑥(𝑛)𝑒
2. Make signal periodic to T seconds (may need window
𝑛=0
function* if there are discontinuities)
3. Expand as a Fourier series
4. Amplitude + phase spectra exist as discrete frequencies and
form the DFT.
FFT has log2(N) *
Periodically
*windowing
N calculations
reduces leakage but
1
extended signal
𝑓𝑠 =
worsens resolution
NO overlapping
𝜏
between adjacent
bands
1. Periodic
Realities:
All signals have ∞ BW so aliasing occurs
Non-ideal filters so no sharp cut-offs
discrete signal
Sampled signal
X(f)
anti-aliasing filter (LPF)
– to reduce noise
-(f+2fs) -(2fs –f) -(f+fs)
-(fs –f)
-f
0
f
fs – f
f+fs
2fs –f
f+2fs
Xs(j2πf) consists of X(j2πf)
plus sidebands centred at nfs
A/D
(sampling)
Shannon’s
sampling
theorem:
Analogue
signal
fs ≥ 2fmax
5
𝑇/2
𝑅0 (𝑓) = ∫
Bode plot for
amplitude
response, AdB(ω) =
20𝑙𝑜𝑔(|𝐻(𝑗𝜔)|)
4
𝑠(𝑡)𝑦(𝑡)𝑑𝑡
−𝑇/2
A peak indicates frequency
of wave:
3
2
1
𝑅𝑒
𝑓𝑠
3. Transfer function, H(z)
End with y[n]
6
Transfer function, H(s)
Frequency response, H(jω)
Amplitude response, |H(jω)| or |H(ejωT)|
𝑖𝑚𝑔
Phase response, ∠𝐻(𝑗𝜔) = 𝑡𝑎𝑛−1 ( )
R0
Zero close to unit circle, Zero on unit For stability, all poles must
Finite Impulse Response
|H(ejωT)| is at a minimum
circle 
be inside unit circle (⨂
filter if just zeros on plane
jωT
|H(e )|=0 at
coincident pole-zero pair
Multiple
(and ole at origin) – linear
notch filter
corresponding
cancels each other out)
phase response
frequency
IIR if there are poles in circle (sharp amplitude
Pole close to unit circle,
response if close to circle)
|H(ejωT)| becomes very large
If R0 > 0, then test wave is
synchronised with original signal
Dirac delta properties:
0
-1
-2
-10
∞
-5
0
Frequency (Hz)
5
10
𝐴𝑇
𝑅0
=
−∞
2
∞
Correlation coefficient,
2
𝑇/2
∫ 𝑔(𝑡)𝛿(𝑡 − 𝑎)𝑑𝑡 = 𝑔(𝑎)
𝑤𝑖𝑑𝑡ℎ =
2𝜋
−∞
𝑇
𝑋𝑅 (𝑓) = ∫ 𝑥(𝑡) cos(2𝜋𝑓𝑡) 𝑑𝑡
𝜔0 =
Dirac delta function when T=∞
−𝑇/2
𝑇
𝑇/2
To find parameters of a wave, s(t):
Area of spike = 1 (unity)
R0 = s(t)y(t)
𝑋𝐼 (𝑓) = ∫ 𝑥(𝑡) sin(2𝜋𝑓𝑡) 𝑑𝑡
To get original signal back:
All waves have:
−𝑇/2
test cosine
∞
Amplitude spectrum
𝑥(𝑡) = ∫ 𝑋(𝑗2𝜋𝑓)𝑒𝑗2𝜋𝑓𝑡 ⁡𝑑𝑓
𝐴(𝑓)
= √𝑋𝑅 2 + 𝑋𝐼 2
Phase spectrum
−∞
Continuous Fourier
−1 ∞
s(t) = A cos(2πft + ϕ)
2
transform:
𝑥(𝑡) =
∫ 𝑋(𝑗𝜔)𝑒𝑗𝜔𝑡 ⁡𝑑𝜔
𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 = 𝐴(𝑓𝑚𝑎𝑥 )
2𝜋 −∞
𝑇
A(f)=|X(j2πf)|
𝑋𝐼 (𝑓max )
ϕ(f) = ∠X(j2πf)
𝑃ℎ𝑎𝑠𝑒 = 𝑡𝑎𝑛−1 (−
)
XR(f)
𝑋𝑅 (𝑓𝑚𝑎𝑥 )
∫ 𝛿(𝑥)𝑑𝑥 = 1
𝑚𝑎𝑥
∞
( f )
𝑋(𝑗𝜔) = ∫ 𝑥(𝑡)𝑒 −𝑗𝜔𝑡 ⁡𝑑𝑡
XI(f)
X(f)
i.e. 𝑠(𝑡) = 3cos⁡(2𝜋(1.5)𝑡 + 1.6)
Ideal ampliture spectra Ideal phase spectra
−∞
2
4
∞
0
Fourier transform: F(s = jω)
{replace s with jω}
3
A m plitude
𝐹(𝑠) = ∫ 𝑓(𝑡)𝑒 −𝑠𝑡 𝑑𝑡
Phase (radians)
3.5
2.5
2
1.5
1.5
1
0.5
1
0.5
0
0
0
0.5
1
Frequency (Hz)
1.5
2
0
0.5
1
Frequency (Hz)
1.5
2
Parsevals theorem:
To solve:
Energy associated with the frequency band ω1 ≤ |ω| ≤ ω2
1. Laplace table
1 𝜔2
2. Fourier (s=jω) or (s=j2πt in terms of
𝐸(𝜔1 , 𝜔2 ) = ∫ |𝐹(𝑗𝜔)|2 𝑑𝑤
𝜋
cyclical frequency)
𝜔1
complex
|𝐹(𝑗𝜔)|2 = 𝐹(𝑗𝜔). 𝐹 ∗ (𝑗𝜔)
3. Amplitude spectrum, A(f) = |F(j2πft)|
conjugate
∞
𝑥(𝑡) = 𝑎𝑛 + ∑(𝑎𝑛 cos(𝑛𝜔0 𝑡) + 𝑏𝑛 sin⁡(𝑛𝜔0 𝑡))⁡
|𝐻(𝑗𝜔𝑐 )| =
𝐻𝑚𝑎𝑥
√2
𝑛=1
Contribution to total power
from N harmonics:
sine (odd period
function)
𝑁
Case 1:
H(s) = s
H(jω) = jω
|H(jω)| = ω
𝐴𝑑𝐵 (𝜔) = 20log⁡(𝜔)
𝑃𝑁 = 𝑎0
2
𝑎𝑛 2 + 𝑏𝑛 2
+∑
2
𝑛=1
𝑇/2
1
= ∫ [𝑓(𝑡)]2 ⁡𝑑𝑡
𝑇 −𝑇/2
a0 = an = 0
4 𝑇/2
𝑏𝑛 = ∫ 𝑥(𝑡) sin(𝑛𝜔0 𝑡) 𝑑𝑡
𝑇 0
Fourier (correlation) coefficents:
1 𝑇/2
1 𝑇
𝑎0 = ∫ 𝑥(𝑡)𝑑𝑡 = ∫ 𝑥(𝑡)𝑑𝑡
𝑓 = 𝑛𝑓0 ⁡
𝑇 −𝑇/2
𝑇 0
2 𝑇
𝑎𝑛 = ∫ 𝑥(𝑡) cos(𝑛𝜔0 𝑡) 𝑑𝑡
a0 : DC term
𝑇 0
n = 1 : fundamental
𝑇
n ≥ 2 : nth harmonic 𝑏 = 2 ∫ 𝑥(𝑡) sin(𝑛𝜔 𝑡) 𝑑𝑡
𝑛
0
𝑇 0
𝑃𝑇𝑂𝑇
Nyquist freq.
Orthogonality doesn’t allow
transmission
all even n = 0
Gain margin, GM = -LGdB
correlation coefficients to
of sampled
Bode
plots
Case 2:
Analogue LPF
Reconstructed
nth harmonic: 𝐴𝑛 cos⁡(𝑛𝜔0 𝑡 + 𝜙𝑛 )
Loop gain, LGdB = 20log|G(jω-180)|
pick up contributions from
signal
H(s) = 1/s
analogue signal
𝑏𝑛
𝑉 (𝑗𝜔)
harmonics other than the nth
cosine (even periodic
Transfer function, 𝐻(𝑗𝜔) = 𝑜
𝜙𝑛 = 𝑡𝑎𝑛−1 (− )
𝐴𝑑𝐵 (𝜔) = −20log⁡(𝜔)
Case 3:
𝑉𝑖 (𝑗𝜔)
harmonic
function) x(t)=x(-t)
𝑎𝑛
GM and PM > 0
Phase margin, PM = ∠𝐺(𝑗𝜔1 ) + 180
H(s) = s + α
for stability
= ∠𝐴𝑑𝐵 (𝜔0 ) + 180
∞
2
When Vi(t) = δ(t) Fourier transform of
𝐴𝑛 = √𝑎𝑛 2 + 𝑏𝑛
|𝐻(𝑗𝜔)| ⁡ = ⁡ √𝜔 2 + 𝛼 2
𝑥(𝑡)
=
𝑎
+
∑
𝑎𝑛 cos⁡(𝑛𝜔0 𝑡)
V
(t)
=
h(t)
is
V
(jω)
=
H(jω)
0
o
o
20log⁡(𝛼),
0≤𝜔≤𝛼
c 
𝐴𝑑𝐵 (𝜔) = {
Overall phase response is sum of phase responses.
𝑛=1
a
=
0
if
+ve
=
-ve
area
δ-function of
0
20 log(𝜔) ,
𝜔>𝛼
For amplitude
Plot H(jω) using linear scale.
height An/2 at
bn = 0
spectrum,
For cut-off frequency:
Case 4:
st
𝑠
𝑇
Butterworth 1 order:
for symmetrical waves, half the
frequency nω0
1
ϕn = 0
𝑠 → for LPF
2
2
Overall bode plot is
𝜔
𝐻(𝑠) ⁡ = ⁡
𝐺
𝑐
graph and multiply by 2
𝑎0 = ∫ 𝑥(𝑡)𝑑𝑡
𝜔
𝑠⁡ + ⁡𝛽
𝐻(𝑠) =
…….
sum of bode plots:
𝑇 0
𝑠 → 𝑐 for HPF
…….
1+𝑠
𝑠
N
2
2
√𝜔 + 𝛼
𝑇/2
|𝐻(𝑗𝜔)|
⁡
=
⁡
2 +𝜔 2
𝑠
4
0
AdB ( )   AdBi ( )
𝑠→
for BPF
20log⁡(𝛽),
0≤𝜔≤𝛽
𝑎𝑛 = ∫ 𝑥(𝑡) cos(𝑛𝜔0 𝑡) 𝑑𝑡
i 1
𝐵𝑠

𝐴𝑑𝐵 (𝜔) = {
𝑇 0
-2
-
0

2
−20 log(𝜔) ,
𝜔>𝛽
where ω02 = 𝜔
̂.̌
𝜔 and B = 𝜔
̂−𝜔
̌
n= 2
n= 1
n=0
n=1
n=2
n
0
0
0
0