Discrete-Time Signals and Systems

Discrete-time signals and systems

1 DISCRETE-TIME DYNAMICAL SYSTEMS

x(t) - G - y(t)

Linear system: Output y(n) is a linear function of the inputs sequence:

∑∞ y(n) = h(k)x(n − k) k=−∞ h(k): impulse response of the system: Impulse

x(0) = 1, x(n) = 0, n ≠ 0 gives y(n) = h(n), all n

2 Some terminology:

• Linear Time Invariant (LTI) System: Impulse response from input x(n − k) to output y(n) does not depend on time n, but only on k

• Causal system: Output y(n) does not depend on future inputs, but only on present and past inputs x(n), x(n − 1),...,

y(n) = h(0)x(n) + h(1)x(n − 1) + ···

(i.e., h(k) = 0, k < 0).

3 Any process which transform a signal {x(n)} to another signal {y(n)} is a discrete-time system (for example, an audio signal is transformed by reﬂections, etc).

Often a (linear) discrete-time system is designed in order to process a signal, for example to remove noise. Then we call such a system a digital ﬁlter.

4 DIGITAL FILTER TYPES

• FIR (Finite Impulse Response) filters - have finite memory; output depends only on a finite number of inputs - modeled by (weighted) moving average models

y(n) = h(0)x(n) + h(1)x(n − 1) + ··· + h(N − 1)x(n + 1 − N)

• IIR (Infinite Impulse Response) filters - have infinite memory; output depends on an infinite number of (past) inputs - modeled by difference equations

y(n) + a1y(n − 1) + ··· + aN y(n − N) =

= b0x(n) + b1x(n − 1) + ··· + bM x(n − M)

5 Calculation of ﬁlter output

FIR ﬁlters: The operation y(n) = h(0)x(n)+h(1)x(n−1)+···+h(N−1)x(n+1−N), n = 0, 1, 2,...,M between the impulse response sequence {h(k)} = {h(0), h(1), . . . , h(N − 1)} and the input sequence {x(n)} = {x(0), x(1), . . . , x(M − 1)} to give the output sequence {y(n)} = {y(0), y(1),...} is called convolution, and is often denoted as

{y(n)} = {h(k)} ∗ {x(n)}

Computational burden: calculation of output sequence requires NM ﬂops

6 Eﬃcient software for DSPs exist

In Matlab: y = conv(h,x) or y = filter(h,1,x) (cf. below)

7 IMPORTANT PROPERTY:

Convolution corresponds to polynomial multiplication:

If we deﬁne the polynomials

Yˆ (z) = y(0) + y(1)z−1 + ··· + y(M − 1)z−M+1 Xˆ(z) = x(0) + x(1)z−1 + ··· + x(M − 1)z−M+1 H(z) = h(0) + h(1)z−1 + ··· + h(N − 1)z−N+1 then we see that ( ) − H(z)Xˆ(z) = h(0)x(0) + h(0)x(1) + h(1)x(0) z 1 + ··· ( ) − + h(0)x(n) + h(1)x(n − 1) + ··· + h(N − 1)x(n + 1 − N) z n + ···

− − = y(0) + y(1)z 1 + ··· + y(M − 1)z M+1 = Yˆ (z)

8 ⇒ the convolution {y(n)} = {h(k)} ∗ {x(n)} can be represented as polynomial multiplication

Yˆ (z) = H(z)Xˆ(z)

Xˆ(z): z-transform of sequence {x(n)} Yˆ (z): z-transform of sequence {y(n)} H(z): transfer function of the system having impulse response h(0), h(1),...

9 Relation between convolution and polynomial multiplication leads to an eﬃcient implementation of convolution

FFT CONVOLUTION (’high-speed convolution’)

The Fourier transform X(k) of {x(n)} is precisely a polynomial in z−1, with z = ej2πk/N :

Recall the Fourier transform of {x(n)}:

X(k) = x(0) + x(1)e−j2πk/N + ··· + x(N − 1)e−j2π(N−1)k/N ( )−1 ( )−(N−1) = x(0) + x(1) ej2πk/N + ··· + x(N − 1) ej2πk/N

Comparison with Xˆ(z) shows that:

j2πk/N X(k) = Xˆ(zk), zk = e

Hence X(k) evaluates Xˆ(z) at N points zk, k = 0, 1,...,N − 1

10 Hence we can compute the convolution as follows:

1. Compute FFT X(k) of sequence {x(n)}. Then X(k) = Xˆ(zk), zk = ej2πk/N .

2. Compute FFT H(k) of sequence {h(n)}. Use zero padding to obtain j2πk/N sequences of the same length. Then H(k) = H(zk), zk = e .

3. Compute Y (k) as the element-wise product Y (k) = H(k)X(k). Then j2πk/N Y (k) = Yˆ (zk), zk = e .

4. Compute inverse Fourier transform {y(n)} of {Y (k)}. Then {y(n)} is the convolution {y(n)} = {h(k)} ∗ {x(n)} TRANSFER FUNCTIONS

x(t) - G - y(t)

In the z-transform domain, we have the simple relation

Yˆ (z) = H(z)Xˆ(z) where H(z) = h(0) + h(1)z−1 + ···

12 GENERAL LINEAR DISCRETE-TIME DYNAMICAL SYSTEMS

General LTI systems can be described by a diﬀerence equation:

y(n) + a1y(n − 1) + ··· + aN y(n − N) =

= b0x(n) + b1x(n − 1) + ··· + bM x(n − M)

Discrete-time counterpart of diﬀerential equations.

Left-hand side is a convolution between the sequence 1, a1, . . . , aN and {y(n)}, or in the z-transform: ( ) −1 −N 1 + a1z + ··· + aN z Yˆ (z)

Right-hand side is a convolution between the sequence b0, . . . , bM and {x(n)}, or in the z-transform: ( ) −1 −M b0 + b1z + ··· + bM z Xˆ(z)

13 ⇒

( ) ( ) −1 −N −1 −M 1 + a1z + ··· + aN z Yˆ (z) = b0 + b1z + ··· + bM z Xˆ(z) or, −1 ··· −M ˆ b0 + b1z + + bM z ˆ Y (z) = −1 −N X(z) 1 + a1z + ··· + aN z

Hence the system has transfer function

−1 ··· −M ˆ ˆ b0 + b1z + + bM z Y (z) = H(z)X(z),H(z) = −1 −N 1 + a1z + ··· + aN z

14 STABILITY Example: First-order system

y(n) + ay(n − 1) = bu(n)

For the input {u(n)} = {1, 0, 0, 0,...} we have:

y(0) = b (assuming y(−1) = 0) y(1) = −ay(0) + bu(1) = −ab y(2) = −ay(1) + bu(2) = a2b . y(n) = (−a)nb

⇒

• y(n) → 0 if as t → ∞ if |a| < 1 – system is stable

• |y(n)| → ∞ as t → ∞ if |a| > 1 – system is unstable

15 Characterization of stability in terms of system transfer function:

b H(z) = 1 − az−1

A value p for which H(p) becomes inﬁnite is called a pole of H(z). Here, p is the solution of 1 − ap−1 = 0 ⇒ p = a. ⇒ System is stable if its pole p = a has absolute value |p| < 1 The results generalizes to higher-order systems: A linear discrete-time dynamical system with transfer function

−1 −M b0 + b1z + ··· + bM z H(z) = −1 −N 1 + a1z + ··· + aN z is stable if and only if all its poles, i.e. the solutions pi of the equation

−1 −N 1 + a1z + ··· + aN z = 0 satisfy |pi| < 1 (i.e., they are inside the unit circle in the complex plane).

16 INTERCONNECTED SYSTEMS Using transfer functions, it is easy to analyze connected systems Example: Series connection.

y1(n) - - - x(n) H1 H2 y(n)

Introducing z-transforms Xˆ(z), Yˆ1(z), Yˆ (z) and transfer functions H1(z),H2(z) gives:

Yˆ (z) = H2(z)Yˆ1(z)

= H2(z)H1(z)Xˆ(z) = H(z)Xˆ(z)

⇒ Series connection has impulse response function {h(k)} with z-transform H(z) = H2(z)H1(z). But this is equal to the convolution {h(k)} = {h2(k)} ∗ {h1(k)}. Hence the impulse response of the series connection is obtained by a convolution of the individual impulse responses

17 Example: Time shift.

Consider a time shift y(n) = x(n − l) Then: Yˆ (z) = z−lXˆ(z)

In other words, if the signal {x(n)} has z-transform Xˆ(z), then the time- shifted signal {x(n − l)} has z-transform z−lXˆ(z). z is a so-called shift operator.

18 Example: Feedback connection.

Determine the output {y(n)} as function of the input {x(n)} in the following feedback connection:

{x(n)} + {e(n)} {y(n)} - i - G - 6−

{r(n)} K where the system G is described by the diﬀerence equation

y(n) − 0.8y(n − 1) = e(n) the K is described by

r(n) − 0.5r(n − 1) = y(n − 1) + 0.1y(n − 2)

19 SOLUTION OF DIFFERENCE EQUATIONS BY THE Z-TRANSFORM In analogy with the the Laplace transform for differential equations, tabulated z-transforms can be applied to solve difference equations as follows: Problem: Determine the solution {y(n)} of the difference equation

y(n) + a1y(n − 1) + ··· + aN y(n − N) =

= b0x(n) + b1x(n − 1) + ··· + bM x(n − M) for a given input sequence {x(n)}. Solution: 1. Write the diﬀerence equation in the z-transform plane:

−1 ··· −M ˆ b0 + b1z + + bM z ˆ Y (z) = −1 −N X(z) 1 + a1z + ··· + aN z

where Xˆ(z) is the z-transform of the input sequence {x(n)}.

2. Solve for Yˆ (z) and ﬁnd the corresponding sequence {y(n)} whose transform Yˆ (z) is.

20 The calculation of Xˆ(z) in stage 1 and of sequence {y(n)} in stage 2 is facilitated by tabulated z-transforms for several common signals (Table 7.1).

Example: z-transform of a signal The exponentially decaying signal { 0, n < 0 x(n) = an, n ≥ 0 has z-transform ∑∞ ∑∞ Xˆ(z) = x(n)z−n = an · z−n n=0 n=0 = 1 + az−1 + a2z−2 + ··· = 1 + az−1 + (az−1)2 + ··· 1 z = = 1 − az−1 z − a

Compare with Table 7.1, line 5, with c = 1, e−α = a.

21 Example: Solution of diﬀerence equation

Consider the diﬀerence equation

y(n) − 0.75y(n − 1) = x(n − 1) with { 0, n < 0 x(n) = (−0.5)n, n ≥ 0

Then: z−1 Yˆ (z) = Xˆ(z) 1 − 0.75z−1 and (cf. above) 1 Xˆ(z) = 1 + 0.5z−1

22 Hence, z−1 1 Yˆ (z) = · 1 − 0.75z−1 1 + 0.5z−1 A partial fraction expansion gives

4/5 4/5 Yˆ (z) = − 1 − 0.75z−1 1 + 0.5z−1 or ( ) 4 z z Yˆ (z) = − 5 z − 0.75 z + 0.5 z { n} ˆ By Table 7.1, z−a is the z-transform of the sequence a . Hence Y (z) is the z-transform of the sequence

4 y(n) = [0.75n − (−0.5)n] 5 which is the solution of the diﬀerence equation.