The Starred Transform and Pulse Transfer Function

6.6 The Starred Transform and Pulse Transfer Function

6.6.1 The Starred Transform

We define the starred transform X ∗ (s) of a signal x(t ) to be the Laplace transform of the impulse-modulated version x ∗ (t ) = x(t)δ T (t ) representing the output signal of an ideal sampler with sampling period T . Using Eq. (2.1.10), we can express the ideal sampler output as

2π x ∗ (t ) = x(t)δ T (t ) =

x (t ) e j kω s t with ω s

= (6.6.1) T

k=−∞

Taking the Laplace transform of the both sides and using the complex translation property (Table B.7(3)), we obtain the bilateral starred transform of x(t ) as

(s + j kω s ) (6.6.2)

k=−∞

k=−∞

298 6 Continuous-Time Systems and Discrete-Time Systems Also, from Appendix III of [P-1], we borrow the following expressions for the

unilateral starred transform:

X ∗ (s) = x (mT ) e −mT s = X[z]| z=e Ts (6.6.3a)

m=0

X ∗ (s) =

Δ x (nT ) e −nT s (6.6.3b) T

where Δx(nT ) = x(nT + ) − x(nT − ) is the amplitude of the discontinuity of x(t ) at t = nT .

The bilateral and unilateral starred transforms have the following properties: < Properties of the starred transform>

1. They are periodic with period of j ω s in s:

X x (nT ) e −nT (s+ jmω s ∗ ) (s + jmω s )= = x (nT ) e −nT s =X ∗ (s) (6.6.4) since e − jn mω s T =e − jn m2π = 1 for any integer m.

2. If they have a pole at s = s 1 , then they will also have poles at s = s 1 + j mω s , m = 0, ±1, ±2, · · · :

X ∗ (s) = T {X(s) + X(s + jω s ) + X(s − jω s ) + X(s + j2ω s ) + X(s − j2ω s )+··· (6.6.5)

6.6.2 The Pulse Transfer Function

Consider the sampled-data systems depicted in Fig. 6.12 where the one has a sam- pled input and the other has a continuous-time input. Under the assumption of zero initial conditions, the output transform of system (a) can be written as

(6.6.6) If x(t ) and y(t ) are continuous at all sampling instants, we can use Eq. (6.6.2) to

Y (s) = G(s)X ∗ (s)

take the starred transform of Eq. (6.6.6) as

T y ∗(t)

x (t )

∗(t)

∗(s) G (s )

y (t )

X (s )

X ∗(s)

Y (s ) T

y ∗(t) Fig. 6.12 Sampled–data

y (t ) systems

x (t )

Y ∗(s) G (s )

X (s )

Y (s )

6.6 The Starred Transform and Pulse Transfer Function 299

Y ∗ (s) = {G(s)X ∗ (s)} ∗ =

G (s + jkω s )X ∗ (s + jkω s )

T k=−∞

G (s + jkω s )X ∗ (s); Y ∗ (s) = G ∗ (s)X ∗ (s) (6.6.7) T

k=−∞

where G ∗ (s) is called the pulse transfer function. This implies that, when taking the starred transform, we can factor out the existing starred transform. If we replace e Ts by z in the starred transform, we will get the z-transform

(6.6.8) which describes the input-output relationship in the z-domain.

Y [z] = G [z] X [z]

In contrast with this, the sampled output of system (b) in Fig. 6.12 can be expressed as

Y ∗ (s) = {G(s)X(s)} ∗ ∗ (s)X ∗ (s) (6.6.9a) Y

(6.6.9b) This implies that, if the input is applied to a continuous-time system before being

sampled, the input transform can never be factored out to derive the transfer function.

6.6.3 Transfer Function of Cascaded Sampled-Data System

Consider the cascaded sampled-data systems depicted in Fig. 6.13. Each of the three sampled-data systems has the input-output relationship as below:

(a) (6.6.7) Y ∗ (s) = {G 2 (s)V ∗ (s)} ∗ =G 2∗ (s)V ∗ (s) = G 2∗ (s)G 1∗ (s)X ∗ (s)

(6.6.10a) (b) (6.6.7) Y

Y [z] = G 2 [z]G 1 [z]X [z]

∗ (s) = {G 2 (s)G 1 (s)X ∗ (s)} ∗ = {G 2 (s)G 1 (s)} ∗ X ∗ (s) = G 2 G 1∗ (s)X ∗ (s)

(6.6.10b) (c) (6.6.7) Y

Y [z] = G 2 G 1 [z] X [z]

∗ (s) = {G 2 (s)V ∗ (s)} ∗ =G 2∗ (s)V ∗ (s) = G 2∗ (s){G 1 (s)X (s)} ∗

=G 2∗ (s)G 1 X ∗ (s); Y [z] = G 2 [z]G 1 X [z] (6.6.10c)

300 6 Continuous-Time Systems and Discrete-Time Systems T

y ∗(t) x (t )

Y (a)

y (t ) X (s )

∗(s) T

∗(t) G 1 (s )

X ∗(s)

V (s )

V ∗(s)

Y (s )

V (s )=G 1 (s )X ∗(s)

Y (s )=G 2 (s )V∗(s) = G 2 (s ){G 1 (s )X∗(s)}∗ T

y ∗(t) x (t )

y (t ) Y ∗(s) (b)

x ∗(t)

v (t )

X (s ) TG

∗(s) Y (s )

V (s ) = G 1 (s )X ∗(s)

Y (s )=G 2 (s )V (s ) = G 2 (s )G 1 (s )X ∗(s) T

y ∗(t) x (t )

y (t ) Y ∗(s) (c)

V 2 ∗(s) (s )

∗(t )

Y (s )

Y (s )=G 2 (s )V∗(s) = G 2 (s ){G 1 (s )X (s )}∗ Fig. 6.13 Cascaded sampled–data systems

V (s )=G 1 (s )X (s )

6.6.4 Transfer Function of System in A/D-G[z]-D/A Structure

Consider the sampled-data system containing an A/D-G[z]-D/A structure or a S/H device depicted in Fig. 6.14. Both of the two systems (a) and (b) have the same transfer function

Y [z]

X [z] =G ho G 2 [z]G 1 [z]

1 [z]X [z] ; V [e ]=G 1 [e ]X [e ]; V ∗ (s) = G 1∗ (s)X ∗ (s)

1∗ (s)X ∗ (s)

1−e −T s

(s) = G 2 (s) ¯ V (s) =

G 2 (s)G (s)X

[z]X [z] = G

z.o.h.

Y (s ) Fig. 6.14 Sampled–data system containing an A/D-G[z]-D/A structure or a S/H device

X (s ) ho Τ X (s ) V (s )

V (s )

Problems 301 where

−T s G 2 G (s) ho G 2 [z] = Z (1 − e ) = (1 − z −1 ) Z

G 2 (s) :

the z.o.h. equivalent of G 2 (s)

(cf.) Note that the DAC usually has a data-hold device at its output, which allows us to model it as a zero-order-hold device.

Problems

6.1 Z.O.H. Equivalent (a) Show that the zero-order-hold equivalent of G(s) = 1/s(s + 1) with

sampling period T is

1 (T − 1 + e −T )z + 1 − e −T −Te −T

G (s) = →G s zoh [z] = (s + 1)

(z − 1)(z − e −T ) (P6.1.1) (b) Show that the zero-order-hold equivalent of G(s) = 2/(s 2 + 2s + 2) with

sampling period T is

G (s) =

(s + 1) +1 (1 − e G −T (cos T + sin T ))z + e −2T −e −T (cos T − sin T )

zoh [z] = z 2 − 2z e −T cos T + e −2T

(P6.1.2)

(c) Use the following MATLAB statements to find the discrete-time equiva- lent of analog system (b) through the BLT (bilinear transformation) with √

sampling period T = 0.1 [s] and critical frequency ω p =

2 [rad/s]. >>B=2; A=[1 2 2]; T=0.1; wp=sqrt(2);

>>GAs= tf(B,A); >>Gz BLT prewarp= c2d(GAs,T,’prewarp’,wp); >>[Bd BLT p,Ad BLT p]= tfdata(Gz BLT prewarp,’v’)

Also support the above results of (a) and (b) by completing and running the following MATLAB program “sig06p 01.m”, which computes the z.o.h. equivalents of analog system (a) and (b) with sampling period T = 0.1 [s].

302 6 Continuous-Time Systems and Discrete-Time Systems

%sig06p 01.m clear, clf B=1; A=[1 1 0]; T=0.1; e T= exp(-T); e 2T= exp(-2*T); GAs= tf(B,A); % Analog transfer function Gz zoh= c2d(GAs,T); % z.o.h. equivalent [Bd zoh,Ad zoh]= tfdata(Gz zoh,’v’) Bd= [(T-1+e T) 1-e T-T*e T], Ad= [1 -1-e T e T] % (P6.1.1)

(d) Referring to the Simulink block diagram of Fig. P6.1, perform the Simulink simulation to obtain the step responses of the analog system (b) and its two discrete-time equivalents, one through z.o.h. and one through BLT.

2 s 2 + 2s + 2

Transfer Fcn [Simulink/Continuous]

Bd_zoh(z) Ad_zoh(z)

Scope [Simulink/Sources]

Step

Discrete Filter

[Simulink/Discrete]

[Simulink/Sinks]

Bd_BLT_p(z) Ad_BLT_p(z)

Discrete Filter 1

Fig. P6.1 Simulink block diagram for Problem 6.1

Fig. P6.2

6.2 Step-Invariant Transformation Consider the system depicted in Fig. P6.2.

(a) Find the step-invariant equivalent G 1 [z] of G 1 (s). (b) Find the step-invariant equivalent G 2 [z] of G 2 (s).

(c) Find the step-invariant equivalent G[z] of G 1 (s)G 2 (s).

(d) Is it true that G[z] = G 1 [z]G 2 [z]?

6.3 Bilinear Transformation without prewarping or with prewarping Consider the second-order analog system whose transfer function is

Problems 303 ω b s

2s

s = 2, ω p = 1 (P6.3.1)

G A (s) = 2 2 = 2 with ω b

+ω b s+ω p

s + 2s + 1

Note that the frequency response G A ( j ω), its peak frequency, and two 3dB- frequencies are

(P6.3.2) + jωω b +ω p 2 (1 − ω 2 ) + j2ω

√ ω p = 1, ω 3B,l =

ω 3B,u =

2 =1+ 2 (P6.3.3) (a) Find the discrete-time equivalent G D [z] using the BLT with sampling

period of T = 1 [s] and no prewarping. Also find the peak frequency Ω p and lower/upper 3dB frequencies Ω 3B,l and Ω 3B,u of the digital frequency

response G D [e jΩ ]. How are they related with ω p ,ω 3B,l , and ω 3B,u ? You

can modify and use the MATLAB program “sig06p 03.m” below.

(b) Find the discrete-time equivalent G D [z] using the BLT with sampling period of T = 1 [s] and prewarping three times, once at ω p = 1, once at

ω 3B,l , and once at ω 3B,u . Also for each G D [z], find the peak frequency Ω p and lower/upper 3dB frequencies Ω 3B,l and Ω 3B,u of the digital frequency response G D [e jΩ ] and tell which frequency is closest to the corresponding

analog frequency in terms of the basic realtionship Ω = ωT between the analog and digital frequencies.

%sig06p 03.m B=[2 0]; A=[1 2 1]; wp=1; w 3dB1= -1+sqrt(2); w 3dB2= 1+sqrt(2); GAs= tf(B,A); T=1; % Analog transfer function and sampling period Gz= c2d(GAs,T,’tustin’); % BLT without prewarping W=[0:0.00001:1]*pi; GDW mag= abs(freqz(BD,AD,W)); [GDW max,i peak]= max(GDW mag); Wp= W(i peak) tmp= abs(GDW mag-GDW max/sqrt(2)); [tmp 3dB1,i 3dB1]= min(tmp(1:i peak)); [tmp 3dB2,i 3dB2]= min(tmp(i peak+1:end)); W 3dB1= W(i 3dB1); W 3dB2= W(i peak+i 3dB2);

6.4 Pole Locations and Time Responses Consider an analog system having the system function G a (s) = 1/(s 2 + 2s + 5) and its two z.o.h. equivalents G zoh1 [z] and G zoh2 [z], each with sampling period T = 0.2 and T = 0.1, respectively.

304 6 Continuous-Time Systems and Discrete-Time Systems

lm{s} lm{z} × 1 2 z – plane

s – plane

2 Re{s}

0 0.5 Re{z}

(a) The pole/zero plot of G a (s)

(b) The pole/zero plot of G zoh1 [z] and G zoh2 [z]

(c) The Simulink model window

(d) The Scope window showing the simulation results Fig. P6.4 Pole/zero plots and Simulink simulation

Problems 305 (a) Note that the pole-zero plot of G a (s) is depicted in Fig. P6.4(a). Refer-

ring to the pole/zero plots shown in Fig. P6.4(b), choose the pole locations (r ∠Ω) of G zoh1 [z] and G zoh2 [z] from 2

n T of the output corresponding to pole location 2 as high as that of the output corresponding to pole location 3 to the sampling period T where ω n is determined from the pole location of

the analog system G a (s). Does it mean that the output corresponding to pole location 2 location 3

(b) Referring to Fig. P6.4(c), perform the Simulink simulation for G a (s),

G zoh1 [z], and G zoh2 [z] to get the simulation results as Fig. P6.4(d) and choose the output waveforms of G a (s), G zoh1 [z] and G zoh2 [z] from 1 and 3

× 2 lm{s}

lm{z}

z – plane

1 s – plane

× × × 2 × Re{z}

1 Re{s}

(a) The pole/zero plot of G a1 (s) and G a2 (s)

(b) The pole/zero plot of G zoh1 [z ] and G zoh2 [z ]

(c) The Simulink model window and the simulation results seen in the Scope window Fig. P6.5 Pole/zero plots and Simulink simulation

6.5 Pole Locations and Time Responses Consider two analog systems, each having the system function G 1 (s) = 1/(s 2 + 2s + 5) and G 2 (s) = 1/(s 2 + 0.4s + 4.04), respectively and their z.o.h. equivalents G zoh1 [z] and G zoh2 [z] with sampling period T = 0.1.

306 6 Continuous-Time Systems and Discrete-Time Systems (a) Note that the pole-zero plots of G 1 (s) and G 2 (s) are depicted in Fig. P6.5(a).

Referring to the pole/zero plots shown in Fig. P6.5(b), choose the pole loca- tions of G zoh1 [z] and G zoh2 [z] from 3 the outputs stemming from the poles 3 closer to the origin than the poles 4

(b) Referring to Fig. P6.5(c), perform the Simulink simulation for G 1 (s), G 2 (s),

G zoh1 [z], and G zoh2 [z] and choose the output waveforms of the four systems from a

6.6 Pole-Zero Mapping (Matched z-Transform) of an Analog System with Delay Using the pole-zero mapping (matched z-transform), find the discrete-time equivalent of an analog system

G (s) =

e −0.95s

(P6.6.1)

s+2

with sampling period T = 1/4 [s].