Digital Controls & Digital Filters Lectures 23 & 24

Non-Recursive Digital Filter Design Digital Controller Design

M.R. Azimi, Professor

Department of Electrical and Computer Engineering Colorado State University

Spring 2017

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Approximation Methods: FIR Digital Filters

FIR filter are described by convolution summation: N−1 y(n) = P h(k)x(n − k) x(n): Input signal y(n): Output signal k=0 h(n): Impulse response (filter coefficients) N: Filter order Advantages:

1 Inherently stable.

2 Linear phase characteristic especially useful in applications where frequency dispersion due to non-linear phase is harmful e.g. speech and audio processing.

3 Quantization eﬀects can be made small.

4 Implementation using parallel/pipeline processors because of inherent concurrency. Disadvantages:

1 For a sharp cut-oﬀ performance a large order ﬁlter is required =⇒ large computation time for implementation.

2 In some cases ”fractional delay” may be needed which can not be realized by physical devices.

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Frequency Response of FIR Filters

The transfer function of a causal FIR ﬁlter is:

N−1 H(z) = P h(n)z−n n=0 Let z = ejΩ (i.e. unit circle) N−1 H(ejΩ) = P h(n)e−jΩn = DTFT {h(n)} n=0 2πk Sampling the frequency scale Ω = N yields the N elements of the DFT of h(n) as:

N−1 − 2πjkn H(k) = P h(n)e N = DFT {h(n)}, k ∈ [0,N − 1] n=0 and N−1 jπjkn 1 P N h(n) = IDFT {H(k)} = N H(k)e , n ∈ [0,N − 1] n=0 Now, consider the following cases.

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Frequency Response of FIR Filters-Cont.

Case 1: Symmetric: h(n) = h(N − 1 − n) and N odd

(N−1) 2 jΩ −jΩ (N−1) H(e ) = e 2 P α(n) cos Ωn n=0 where h (N−1) i N−1 α(n) , 2h 2 − n , n ∈ [1, 2 ] (n−1) α(0) , h 2 (N−1) φ(Ω) = −Ω 2 : linear

dφ(Ω) N−1 Group delay τg = − dΩ = 2 : Integer

Case 2: Symmetric: h(n) = h(N − 1 − n) and N Even

N/2 jΩ −jΩ (N−1) H(e ) = e 2 P β(n) cos [Ω(n − 1/2)] n=1 where, N β(n) , 2h( 2 − n), n ∈ [1,N/2] Note that at Ω = π, H(ejΩ) = 0 i.e HPF can not be approximated.

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Frequency Response of FIR Filters-Cont.

N−1 Case 3: Anti-Symmetric: h(n) = −h(N − 1 − n), h( 2 ) = 0, and N odd

(N−1) 2 jΩ −jΩ (N−1) jπ/2 H(e ) = e 2 e P γ(n) sin Ωn n=1 where h (N−1) i N−1 γ(n) , 2h 2 − n , n ∈ [1, 2 ]

Case 4: Anti-Symmetric: h(n) = −h(N − 1 − n) and N even

N/2 jΩ −jΩ (N−1) jπ/2 H(e ) = e 2 e P δ(n) sin [Ω(n − 1/2)] n=1 where, N δ(n) , 2h( 2 − n), n ∈ [1,N/2] In the anti-symmetric case at Ω = 0,H(ejΩ) = 0. i.e. suitable for approximating such ﬁlters as diﬀerentiators and Hilbert transforms

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Frequency Sampling Method

Design Methods 1 Windowing 2 Frequency sampling method 3 Computer aided design methods (CAD) Frequency Sampling Method jΩ Idea: Given the frequency response of the desired ﬁlter, HDes(e ), sample it to obtain H(k) and then take IDFT of {H(k)} to get {h(n)}.

Let the transfer function of the FIR ﬁlter be : N−1 H(z) = P h(n)z−n n=0 we know that N−1 −j2πkn P N H(k) = H(z)|z= j2πk = h(n)e , k ∈ [0,N − 1] N n=0

N−1 j2πkn 1 P N Additionally: h(n) = IDFT {H(k)} = N H(k)e , n ∈ [0,N − 1] k=0 Then, N−1 N−1 N−1 N−1 j2πkn H(k) j2πk P P N −n P P N −1 n H(z) = H(k)e z = N (e z ) n=0 k=0 k=0 n=0

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Frequency Sampling Method-Cont.

N−1 P n 1−aN Now using a = 1−a , we get , n=0 N−1 −N N−1 P H(k) (1−ej2πkz−N ) (1 − z ) X H(k) N j2πk = j2πk k=0 (1−e N z−1) N (1 − e N z−1) | {z } k=0 H1(z) | {z } H2(z) which can be realized by cascade of a simple FIR (Comb) ﬁlter with an IIR system.

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Frequency Sampling Method

The term (1 − z−N ) represents an FIR ﬁlter with frequency response N h N N i N jΩ −jΩN −jΩ 2 jΩ 2 −jΩ 2 −jΩ 2 ΩN H1(e ) = 1 − e = e e − e = e 2j sin 2 jΩ The plot of |H1(e )| as a function of Ω is shown below.

The IIR part, H2(z), consists of parallel combination of N complex coeﬃcient j2πk 1st order systems with poles at zk = e N on the unit circle. Of course these j2πk poles will be cancelled with the zeros of H1(z) which also occur at zk = e N −N 1/N j2πk since 1 − z = 0 =⇒ z = (1) =⇒ zk = e N , k ∈ [0,N − 1]

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Frequency Sampling Method

In order to have H(z) with real coeﬃcients (physical realization) we require that H(k) = |H(k)|ejθ(k) satisfy:

|H(k)| = |H(N − k)| and θ(k) = −θ(N − k), k ∈ [0,N − 1]

N Now each pair of H(k) and H(N − k), k ∈ [0, 2 ] gives a 2nd order system with real coeﬃcients. i.e: −1 2πk H(k) H(N−k) 2|H(k)|[cos θ(k)−z cos (θ(k)− N )] Hk(z) = + = j2πk j2π(N−k) 1−2 cos 2πk z−1+z−2 1−e N z−1 1−e N z−1 N Note that for N odd there is no frequency sample for k = N/2 thus the corresponding term does not exist.

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Frequency Sampling Method: Design Procedure

jω Given desired frequency response HDes(e ) sample it to yield:

jΩ H(k) = HDes(e )| 2πk , k ∈ [0,N − 1] Ω= N Impose the following conditions on the magnitude |H(k)| and phase θ(k):

(N−1)kπ 1 For linear phase: θ(k) = − N , k ∈ [0,N − 1] 2 For real coeﬃcients: |H(k)| = |H(N − k)| and θ(k) = −θ(N − k)

N Note for N even set H( 2 ) = 0. Example: Design an N = 11 order low-pass FIR filter whose cut-off frequency fc = 2fs/11, where fs is the sampling frequency. Plot the magnitude response of the designed filter. jΩ We start with ideal filter for HDes(e ) and sample it with spacing between samples= Ωs fs 2πN = N . See below figure.

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Frequency Sampling Method: Design Procedure

We then impose the above conditions i.e.

|H(k)| = |H(N − k)| and

(N−1)kπ 10kπ θ(k) = − N = 11 , k ∈ [0, 10] These combined yield,

−j 10kπ e 11 k = 0, 1, 2 and k = 9, 10 H(k) = 0 k ∈ [3, 8]

Then take size N = 11 IDFT of H(k) i.e. h(n) = IDFT {H(k)}11 which gives h(0) = 0.06942, h(1) = −0.05403, h(2) = −0.10945, h(3) = 0.04733, h(4) = 0.31938, h(5) = 0.45455, h(6) = 0.31938, h(7) = 0.04733, h(8) = −0.10945, h(9) = −0.05403, h(10) = 0.06942

Note that these are symmetric due to the linear phase condition imposed. Magnitude response of the designed ﬁlter is shown (thick line) which exhibits a rather large ripple in both passband and stopband. To reduce the ripple eﬀect, one can add a sample in the transition (wider transition region) region, i.e.

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Frequency Sampling Method: Design Procedure

 −j 10kπ  e 11 k = 0, 1, 2 and k = 9, 10 −j 10kπ H(k) = 0.5e 11 k = 3 and k = 8  0 k ∈ [4, 7]

The resulting ﬁlter coeﬃcients are: h(0) = 0.00987, h(1) = 0.02244, h(2) = −0.07168, h(3) = −0.03989, h(4) = 0.30643, h(5) = 0.54545, h(6) = 0.30643, h(7) = −0.03989, h(8) = −0.07168, h(9) = 0.02244, h(10) = 0.00987

The magnitude response is shown (thin line) which shows much less ripple eﬀects at a cost of wider transition region (must increase the order). Note that one can try to optimize the choice of the amplitude of the introduced sample.

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Design of Digital Controller-Classical method

Goal: Design compensators to achieve certain desired characteristics e.g., steady-state accuracy, transient response, relative stability, sensitivity, etc.

1 Steady-State Accuracy: Desired ess ≈ 0 =⇒ poles at z = 1 =⇒ more lags =⇒ less stable

2 Transient Response: Desired: Reduce tr =⇒ increase BW =⇒ less noise immunity

3 Relative Stability: Desired: Small Mp =⇒ increase in tr, slower response

4 Sensitivity: Variation of system parameters wrt change in temperature, humidity, aging, altitude, etc. T ∂T a G(z) Measured by Sα = ∂α . T where T (z) = 1+G(z) and α is a system parameter, e.g.,

T 1 T 1 SG = 1+G(z) or in frequency domain SG = 1+G(ejΩ) which implies,

T jΩ Reduce sensitivity SG =⇒ Increase open loop gain G(e ) =⇒ Instability problems.

M.R. Azimi Digital Control & Digital Filters Non-Recursive Digital Filter Design Digital Controller Design Digital PID Controller

Recall from ECE 411, for PID controller:

Gc(s) = KP + KI /s + KDs

where KP : proportional gain, KI integrator gain, and KD derivative gain. Though bilinear z-mapping can be used for the integrator: T (z+1) 1/s → 2 z−1) it cannot be used for the derivative part. But we can use backward diﬀerence method: dx(t) x(t)−x(t−∆T ) y(t) = dt ≈ ∆T , ∆T = 1, t = nT x(n)−x(n−1) (X(z)−z−1X(z)) Y (z) (z−1) y(n) = T =⇒ Y (z) = T =⇒ X(z) = zT (z−1) Thus, the mapping is s → T z which yields digital PID transfer function: 2 T (z+1) (z−1) a0z +a1z+a2 Gc(z) = KP + KI 2 (z−1) + KD T z = z(z−1) : Second order IIR ﬁlter

where a0 = KP + KI T/2 + KD/T , a1 = −KP + KI T/2 − 2KD/T , a2 = KD/T , b1 = −1, and b2 = 0

Design of PID controllers involves ﬁnding KP , KI , and KD to meet the desired performance criteria.