III. Time Domain Analysis of systems

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1 1 III. Time Domain Analysis of systems Here, we adapt properties of continuous time systems to discrete time systems Section , pp System Notation y(n) = T[ x(n) ] A. Types of Systems Memoryless (instantaneous) Systems y(n) = function of x(n), not x(n+k) for k 0. Example Are these systems memoryless? (1) T[ x(n) ] = sin (x(n))

2 2 (2) Accumulator: T[ x(n) ] = x(k) Linear Systems. Definition: T[ax 1 (n)+bx 2 (n)] = at[x 1 (n)] + bt[x 2 (n)] Test for linearity? Manipulate LHS and compare to RHS. Example T[x(n)] = x(k) T[a x 1 (n)+b x 2 (n)] = a x 1 (k)+b x 2 (k) = a x 1 (k) + b x 2 (k)

3 3 a T [ x 1 (n) ] + b T[ x 2 (n) ] = same Example Given : T[ x(n) ] = e x(n) Is this system memoryless? Is it linear? a x1(n) + b x2(n) T[ax 1 (n) + bx 2 (n)] = e = (e x1(n) ) a (e x2(n) ) b Now, at[x 1 (n)] + bt[x 2 (n)] = ae x1(n) + be x2(n) The system is nonlinear.

4 4 Determining Linearity by Inspection In the description of a linear system, every term has an input to the first power or an output to the first power, but not both. Time-Invariant Systems Definition: a System T[x(n)] is time invariant if y(n) = T[x(n)] T[x(n-k)] =y(n-k) Test : (1) Find y 1 (n) = T[x 1 (n)] and replace x 1 (n) by x(n-k) (subtract k from argument of x 1 ) (2) Find y 2 (n) = T[x(n)] (3) System is time invariant if y 2 (n-k) = y 1 (n) ( For y 2 (n), perform n n-k )

5 5 Determining Time Invariance by Inspection In the description of a time invariant system, (a) Every coefficient of an input or an output should not be a function of time. (b) In the argument of x(), n should not be multiplied by anything except 1. Example Is the system described by T[x(n)] = n 2 x(n-1) = y(n) time invariant? y 1 (n) = n 2 x 1 (n-1) = n 2 x(n-k-1) y 2 (n) = n 2 x(n-1) y 2 (n-k) = (n-k) 2 x(n-k-1) The system is not time invariant. It is time variant.

6 6 Example Is the compressor, y(n) = x(mn), time invariant? Solution 1: y 1 (n) = x 1 (Mn) = x(m n-k)] y 2 (n) = x(mn) y 2 (n-k) = x(m(n-k)) Solution 2:

7 7 Stability (1) BIBO Stability A system is BIBO stable iff a bounded input always produces a bounded output. Sketch of Bounded Input Sketch of Bounded Output

8 8 Theorem: A linear time invariant system is stable iff S <, where S = h(k) Proof: Case 1 Assume h(k) <, then y(n) = h(k) x(n-k) h(k) x(n-k) M h(k) < where x(n) M < and M is upper bound on x(n). system is stable. Case 2 Assume system is stable y(n) = h(k) x(n-k)

9 9 Largest y(0) occurs for x(-k) = M h(k) / h(k) so h(k) y(0) = h(k) M h(k) = M h(k) < Since system is stable, y(n) also has M h(k) < as maximum value by time invariance. h(k) < QED

10 10 Testing System for stability (1) First determine whether or not the system is linear time invariant (LTI). s; The system is LTI if T[x(n)] = h(k) x(n-k) where (a) k 1 and k 2 can be made to be constants, independent of n. (b) h(k) is not a function of n or x(n). (2) If the system is LTI, then it is stable iff h(k) <. (3)If the system is not LTI, it is stable if max y(n) < B for all sequences {x(n)}

11 11 such that x(n) < M. Check this by finding B as a function of M. This may be done by finding a sequence {x(n)} which maximizes y(n) or by bounding y(n) by some other method. Example Determine the stability of a system described by T[ x(n)] = x(n) + 2 x(n-1) = h(k) x(n-k) where h(0) = 1, h(1) = 2, S = h(0) + h(1) < BIBO Stable

12 12 Example Determine the stability of a system described by T[x(n)] = a k x(n-k) h(n) = a n u(n) so the system is LTI. S = h(n) = a n = a n 1 = if a < 1 (1 - a ) Stable if a < 1, unstable otherwise.

13 13 Example Determine the stability of a system described by T[x(n)] = n x(n) Test for Time Invariance y 2 (n) = n x(n) y 2 (n-k) = (n-k) x(n-k) y 1 (n) = n x 1 (n) = n x(n-k) y 1 (n) y 2 (n-k) so not time invariant. Alternately, by inspection, y(n) = h(k) x(n-k) h(0) = n Not LTI function of time.

14 14 Test For Stability y(n) = n x(n), x(n) < M y(n) = n x(n) M n y(n) approaches, even if x(n) M, so y(n) is unbounded and the system is not stable. Example Determine the stability of a system described by T[x(n)] = x(n) (1+n 2 ) By inspection, the system is time variant.

15 15 x(n) y(n) = M. (1+n 2 ) so the output is bounded. System is stable. Example Determine the stability of a system described by T[x(n)] = x(n-k) (1+(k-n) 2 ) By inspection, the system is time variant. Stable? x(n-k) y(n) = (1+(k-n) 2 )

16 16 x(n-k) (1+(k-n) 2 ) M 1 (1+(k-n) 2 ) [1+2 1/n 2 ] M < Stable. Causality A causal system is one for which y(n) is a function of x(n-k) where k 0. General Definition: A LTI system is causal iff h(n) = 0 for n < 0.

17 17 Test for Causality (1) Determine whether or not the system is LTI. (2) If it is LTI, then the system is causal if h(n) = 0 for n < 0. (3) Otherwise, show that y(n) is a function of x(n-m) for m 0 and not a function of x(n+m) for m > 0. Example T[x(n)] = h(k) x(n-k) Causal since it is a function of x(n-k), k 0. Also h(n) = α n u(n) = 0 for n < 0, assuming LTI. Example T[x(n)] = x 2 (n+1) + x 2 (n-1) + x(n) = y(n) System is not causal.

18 18 Example Ideal Time Delay system y(n) = x(n-n d ) n d > 0 Memoryless? No. Linear? x(t) * δ(t-t o ) = x(t-t o ) Time invariant? Yes. y 1 (n) = x 1 (n-n d ) = x(n-n d -k) y 2 (n) = x(n-n d ) y 2 (n-k) = x(n-n d -k) Causal? Yes, if n d 0 Stable? x(n) M y(n) M BIBO stable for all n. S = h(n) = 1

19 19 Example Moving Average y(n) = 1 (M 1 + M 2 + 1) x(n-k) M 1 0, M 2 0 y(n) = 1 (M 1 + M 2 + 1) [δ(n-m 2 ) + δ(n-m 2 +1) +. δ(n+m 1 )]*x(n) Memoryless; No. Linear; Yes. Time Invariant; Yes. Causal; Maybe if Stable; Yes.

20 20 S = h(n) = (M 1 +M 2 +1) 1 (M 1 + M 2 + 1) = 1

21 21 Example Systems System : Ideal Delay. h(n) : δ(n-n d ) Characteristics: Stable, Causal, FIR. System: Moving Average. h(n) : δ(n-k) M 1 +M 2 +1 Characteristics : Stable, FIR, Maybe noncausal. System : Accumulator. h(n) : u(n). y(n) = u(k) x(n-k) = x(n-k) Characteristics: IIR, Causal, unstable.

22 22 System: Forward Difference. h(n): δ(n+1) - δ(n) Characteristics: non-causal, FIR, Anti- Causal, stable. System: Backward Difference. h(n) : δ(n) - δ(n-1) Characteristics: Stable, Causal, FIR. System: Causal Exponential. h(n): a n u(n) Characteristics : Causal, Stable iff a < 1, IIR. S = a n = 1 (1- a ) iff a < 1, else S =.

23 23 Example Inverse system example (Deconvolution of motion blur of Neptune?) h(n) = u(n), accumulator. h(n) = backward difference, δ(n) - δ(n-1) h(n) * h(n) = u(k) [ δ(n-k) - δ(n-k-1)] = u(n) u(n-1) = δ(n) x(n) h(n) h(n) y(n) x(n) δ(n) y(n) = x(n) Problems; h(n) is unstable, can get infinite signals here. Find Inverse systems using Z-transforms.

24 24 C. Linear Constant Coefficient Difference Equations 1. Preliminary Examples Convolution is the most general description of LTI systems, but speed is important and computing power is limited. People use an FIR filter with few coefficients or recursive solution to difference equations. y(n) = h(k) x(n-k) Here we have 1+2M multiplications per output sample. This is an FIR filter.

25 25 Example Let h(n) = a n u(n). Find the number of multiplies per output sample for three different solutions. Solution 1 y(n) = a k x(n-k) = a k x(n-k) This requires n+1 multiplications per output sample. The work grows as time progresses. y(0) = x(0). y(1) = x(1) + a x(0), y(2) = x(2) + ax(1) + a 2 x(0) y(3) = x(3) + a x(2) + a 2 x(1) + a 3 x(0) etc. Solution 2: FIR y(n) = a k x(n-k)

26 26 Solution 3 Consider the linear constant coefficient difference equation: y(n) = x(n) + a y(n-1) y(0) = x(0)+a y(-1), y(1) = x(1) + a y(0) = x(1) + a x(0) y(2) = x(2)+a y(1) = x(2) + a x(1) + a 2 x(0) y(n) = a k x(n-k) and y(n) = x(n) + a y(n-1) give identical results.

27 27 The third solution has one multiplication per output point versus (n+1) for the first solution. The third solution which is much better, is recursive. Example Causal Moving Average y(n) = 1 (1+M 2 ) x(n-k) M 2 additions for 1+M 2 terms. But 1 y(n+1) = 1+M 2 x(n+1-k) = y(n) + [x(n+1) x(n-m 2 )] / (1+M 2 ) y(n) = y(n-1) + C (x(n) - x(n-1-m 2 ) )

28 28 where 1 C = 1+M 2 2 adds and one multiply. 2. Particular and Homogenous Solutions In continuous time, linear constantcoefficient differential equations have a homogeneous solution and a particular solution. The same thing happens in discrete time with linear constantcoefficient difference equations: a k y(n-k) = b k x(n-k) y(n) = y p (n) + y h (n) y h (n) is any solution for x(n) = 0 and

29 29 y p (n) is a solution due to x(n), assuming y(n) is initially zero. a k y h (n-k) = 0 y h (n) = A m Z m n where Z m are roots of the polynomial. a k Z -k = 0. Example Given the difference equation y(n) = x(n) + 5/6 y(n-1) 1/6y(n-2) where x(n) and y(n) satisfy initial rest conditions at time 0. (y(n) = x(n) = 0 for n < 0) get output y(n). (1- (5/6)Z -1 + Z -2 /6)

30 30 = ( Z 2 (5/6) Z + 1/6) = (Z- 1/2)(Z- 1/3) y h (n) = A 1 (1/2) n + A 2 (1/3) n homogenous solution Example Given y h (0) = 2 and y h (1) = 3 in the previous example, find A 1 and A 2 y h (0) = A 1 + A 2 = 2 y h (1) = (1/2) A 1 + (1/3) A 2 = 3 use diff. Eqn.

31 31 D. Convolution Examples Ex x(n) = 2 δ(n) - δ(n-1) h(n) =- δ(n) + 2 δ(n-1) + δ(n-2) y(n) = h(k) x(n-k) = [ - δ(k) + 2 δ(k-1) + δ(k-2)] [ 2δ(n-k) - δ(n-k-1)] k = 0 k =1 k = 2-2δ(n) + δ(n-1) 4 δ(n-1) 2 δ(n-2) 2 δ(n-2) - δ(n-3) y(n) = -2δ(n) + 5 δ(n-1) - δ(n-3)

32 32 Ex. h(n) = α n u(n), x(n)= β n u(n) y(n)=? Ex h(n) = α n 0 n N-1 0 elsewhere x(n) = β n-n0 n 0 n 0 n < n 0

33 33 Ex h(n) non-zero for N 0 n N 1 x(n) non-zero for N 2 n N 3 y(n) non-zero for N 4 n N 5 Given N 0, N 1, N 2, N 3, find N 4 and N 5. y(n) = h(k) x(n-k) (1) Put limits on k N 0 k N 1 N 2 n-k N 3 (2) Put limits on n as a function of k.

34 34 N 2 + k n N 3 + k (3) Now let k take on its extreme values. (N 2 + N 0 ) n (N 3 + N 0 ) (N 2 + N 1 ) n (N 3 + N 1 ) N 4 = min { ( N 2 + N 0 ), ( N 2 + N 1 ) } N 5 = max { (N 3 + N 0 ), (N 3 + N 1 ) }

35 35 Ex h(n) = x(n) = u(n) Ex h(n) = u(n+2) u(n-3), x(n) = u(n+1) u(n-6)

36 36 Ex h(n) = cos(w 1 n), x(n) = u(n) Ex h(n) = a n u(n), x(n) = u(n)

37 37 E. Advanced Topic: Fitting a Polynomial to a Signal Given x(n) for 0 n N-1, approximate x(n) as x( n) M 1 k = 0 b n k k where n can be a real variable on the right hand side. Define an error function as N 1 M 1 k E = [ x( n) b n ] n= 0 k = 0 k 2 Taking derivatives as E a m N 1 M 1 k m = 2 [ x( n) b n ] n = 0, n= 0 k = 0 k

38 38 we get M 1 N 1 N 1 k m m b + k n = x n n k = 0 n= 0 n= 0 M 1 k = 0 a( m, k) b = c( m), k ( ), A b = c a(m, k) = a(m + k) where a(m) N 1 c(m) = x( n) n n= 0 m = N 1 n= 0 n m Given A and c, we can solve for b

39 39 Feature Vector Viewpoint Let the vector X contain a(m) for 0 m 2M-2, and c(m) for 0 m M-1. Therefore the dimension of X is N F = 3(M-1). Example Applications (1) Integrate x(t) from t = 0 to t = N-1 (2) How many times does x(t) go to zero for t between 0 and N-1? (3) Find the number of maxima, minima, and points of inflection for x(t) with t between 0 and N-1.

VU Signal and Image Processing

052600 VU Signal and Image Processing Torsten Möller + Hrvoje Bogunović + Raphael Sahann torsten.moeller@univie.ac.at hrvoje.bogunovic@meduniwien.ac.at raphael.sahann@univie.ac.at vda.cs.univie.ac.at/teaching/sip/18s/