x[n] = x a (nt ) x a (t)e jωt dt while the discrete time signal x[n] has the discrete-time Fourier transform x[n]e jωn

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1 Sampling Let x a (t) be a continuous time signal. The signal is sampled by taking the signal value at intervals of time T to get The signal x(t) has a Fourier transform x[n] = x a (nt ) X a (Ω) = x a (t)e jωt dt while the discrete time signal x[n] has the discrete-time Fourier transform X[ω] = n= x[n]e jωn The question is: what is the relationship between these two Fourier transforms? The following theorem gives the answer Theorem. Let x[n] = x a (nt ) be a sampled version of the continuous time signal x a (t). Then the relationship between the Fourier transforms is (assuming that all signals have nice Fourier transforms) X[ω] = 1 ( ) ω + k2π X a (1) T T k= Proof. First notice that X[ω] is periodic with period 2π. It is therefore the Discrete Time Fourier Transform (DTFT) of some sequence x[n], and we can calculate the inverse Fourier transform as follows x[n] = 1 2π 2π = 1 2π = 1 1 2π T = 1 1 2π T 0 2π 0 = 1 1 2π T = 1 1 2π T X[ω]e jωn dω 1 T k= 2π k= 0 (k+1)2π k= k2π X a ( ω T = x a (nt ) = x[n] ( ω + k2π X a T X a ( ω + k2π T ( ω X a T ) e jωn dω X a (ω) e jωt n d(ωt ) ) e jωn dω ) e jωn dω ) e jωn dω Since the inverse Fourier transform of X[ω] is x[n], X[ω] is indeed the Fourier transform of x[n]. 1

2 It is often more convenient to use the continuous time frequency Ω instead of the discrete time frequency ω. Define Then (1) becomes X p (Ω) = X[ω] ω=ωt X p (Ω) = 1 T Ω s = 2π T k= X a (Ω kω s ) (2) Let x[n] be a discrete time signal, and y(t) a continuous time signal. Then we define the convolution between x[n] and y(t) with period T by z(t) = x[n] T y(t) def = n= Theorem. The Fourier transform Z( Ω) of z(t) is given by x[n]y(t nt ) Z(Ω) = X p (Ω)Y (Ω), X p (Ω) = X[ω] ω=ωt X[ω] = x[n]e jωn Y (Ω) = n= y(t)e jωt dt (3) Proof. Exercise. 1 X(ω) 1/Τ X Τ (ω) Τ Η(ω) X(ω)=Η(ω)X Τ (ω) Figure 1: Sampling Reconstruction 2

3 This result can be used for perfect reconstruction of a bandlimited signal. Suppose that x a (t) has maximum frequency Ω m, that is X a (Ω) = 0 for Ω > Ω m. In that case (see figure), X(Ω) = H(Ω)X a (Ω) with Then according to the above we get H(Ω) = { T Ω Ωs 2 0 otherwise Here (from a CTFT table) So, x(t) can be reconstructed from x(t) = x[n] T h(t) = x(t) = n= ( ) t h(t) = sinc T sinc(x) = sin(πx) πx n= x[n]h(t nt ) ( ) t x[n]sinc T n 3

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5 Impulse Response for Finite Difference Equations Consider the finite difference equation N M d k y[n k] = p j x[n j]; y[ 1] = y[ 2] = y[ N] = 0 k=0 j=0 The procedure for finding the impulse response is 1. Find the homogenous solution. For simplicity of notation, assume that all roots are different y h [n] = N α i λ n i 2. Find the particular (transient) solution by manual calculation, using initial conditions, y[n] = p n d 0 N k=1 d k d 0 y[n k], n = 0, 1,... M 3. Find the coefficients α i by using y[m], y[m 1],... y[m N + 1] as initial conditions 4. The impulse response is α 1 λ1 M N+1 + α 2 λ M N α N λ M N+1 N = y[m N + 1] α 1 λ1 M N+2 + α 2 λ M N α N λ M N+2 N = y[m N + 2] α 1 λ M 1 + α 2 λ M α N λ M N = y[m] h[n] = y[n], n = 0, 1,..., M N (from step 2) N h[n] = α i λ n i, n = M N + 1,..., (α i found in step 3). 5

6 Frequency Analysis of Continuous Time signals Consider a continuous-time signal x(t), which is a superposition of sinusoidal signals x(t) = K A i cos(ω i t + θ i ) We would like to use the DFT (or, rather, the FFT) to determine the frequencies ω i of this signal. The Fourier transform of x(t) is X c (ω) = x(t)e jωt dt = K K πa i e jθ i δ(ω ω i ) + πa i e jθ i δ(ω + ω i ) So that the frequencies ω i can be found from the peaks of the Fourier transform. The signal is sampled with a sampling interval T to give x[n], x[n] = K A i cos(ω i T n + θ i ) π Provided the sampling frequency is high enough to avoid aliasing, i.e., T < max i {ω i }, the DTFT of x[n] is given by K K X(ω) = πa i e jθ i δ(ω ω i T ) + πa i e jθ i δ(ω + ω i T ) Thus, the frequencies ω i can be found from the sampled signal by performing a DTFT, and the relationship between the DTFT and the Fourier transform is X c (ω) = X(ω/T ) In praxis we only have a finite number of samples, x[n], 0 n N 1, and the DFT is used to find the spectrum. Define { x[n] 0 n N 1 x N [n] = 0 otherwise Then we can write The DFT of x N [n] is then given by where X N (ω) is the DTFT ofx N [n] x N [n] = x[n]w N [n] { 1 0 n N 1 w N [n] = 0 otherwise X[k] = DF T (x N [n]) = X N (ω) ω=2πk/n, 0 k N 1 π X N (ω) = DT F T (x[n]w N [n]) = 1 X(ψ)W N (ω ψ)dψ 2π π K K 1 = 2 A ie jθ i 1 W N (ω ω i T ) + 2 A ie jθ i W N (ω + ω i T ) 6

7 and W N (ω) is the DTFT of w N [n]: W N (ω) = N 1 n=0 { N ω = 0 1 e jωn = otherwise 1 exp( jωn) 1 exp( jω) Thus, the DFT does not directly give the peaks of the DTFT, but instead the function W N (ω). Figure 2: Frequency analysis of the signal x(t) = cos(2π395t) However, this can usually be used for frequency analysis if N is large enough, as will be seen from some examples. To be able to interpret a plot we need to have the right frequency range. The DFT gives X N (ω) at the frequencies 0, 2π 1/N, 2π 2/N,..., 2π (N 1)/N, i.e., in a frequency range of approximately 0 2π. However, we are interested in the frequencies in the range π π. This can be fixed using the periodicity of X N (ω) by setting X N (ω) ω=2πk/n = X[k + N], N 2 k 1 X N (ω) ω=2πk/n = X[k], 0 k N 2 1 The easiest way to implement this is through the Matlab function fftshift. There is still the problem that we only have a frequency resolution of 2π 1/N, i.e., the step size in the frequency domain. If we are interested in a frequency that is not a multiple of 2π 1/N, the DFT does not give us this. The frequency resolution can be increased by zero-padding x N [n]. Define the length M > N signal x N,M [n], 0 n M 1 by { x[n] 0 n N 1 x N,M [n] = 0 N n M 1 7

8 Then the DFT of x N,M [n] is X N,M [k] = DF T (x N,M [n]) = X N (ω) ω=2πk/m, 0 k M 1 So now the output frequencies are 0, 2π 1/M, 2π 2/M,..., 2π (M 1)/M, i.e., the frequency resolution is 2π 1/M < 2π 1/N. The following Matlab program integrates this approach to frequency analysis Nf=zp length ( x ) ; % zp i s the zero padding f a c t o r ( an i n t e g e r ) y=f f t ( x, Nf ) ; % appends a u t o m a t i c a l l y Nf length ( x ) z e r o s y=f f t s h i f t ( y ) ; %s h i f t the two h a l v e s o f the FFT f r e q =(0: Nf 1)/Nf/T f s / 2 ; % c a l c u l a t e frequency range stem ( freq, abs ( y ) ) ; % p l o t the magnitude Example 1. Let x(t) = cos(2π395t) and let the sampling frequency be 1000Hz. We use N = 64 samples of the signal for the DFT. The result can be seen in Figure. The reason the plot is not a single peak is due to the convolution with the function W N (ω). Figure 3: Frequency analysis of the signalx(t) = cos(2π388t) + cos(2π400t) Example 2. Let x(t) = cos(2π388t) + cos(2π400t) and let the sampling frequency be 1000Hz. We use N=64 samples of the signal for the DFT. The result can be seen in Figure. As can be seen, without zeropadding, the two sinusoids cannot be distinguished it seems like there is only one peak. With zeropadding it can be seen that there are two peaks. However, it is not always possible to distinguish the peaks, and DFT (or FFT) is therefore not the best tool for frequency analysis of 8

9 signals with multiple frequencies. For more detail and advanced methods, see chapter 11.2 in the textbook and [1][2]. References 1. S.M. Kay, Modern Spectral Estimation: Theory and application, Prentice-Hall, S.M. Kay and S.M. Marple, Jr., Spectrum analysis-a modern perspective, Proceedings of the IEEE, vol.69, no.11, p ,

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12 Inversion of the Z-transform G(z) = P (z) D(z) = p 0 + p 1 z p M z M d 0 + d 1 z d N z N 1. If M N rewrite as G(z) = M N = p 0 + p 1 z p M z M N n=1 (1 λ nz 1 ) η i z i + P 1(z) D(z) }{{} G 1 (z) where order(p 1 (z)) <order(d(z)). P 1 (z) and η 1,..., η M N is found by long division (see section 6.4.5). 2. Do a partial fraction expansion of G 1 (z). There are several cases a) Only simple real poles (all λ i different). Write G 1 (z) as G 1 (z) = N ρ i 1 λ i z 1 The coefficients ρ i can be found by ρ i = (1 λ i z 1 )G 1 (z) z=λi b) One or more (simple) complex poles. If the z transform has real coefficients, complex poles come in complex conjugate pairs. For example, suppose λ 1 is complex. Then there is some other pole λ i so that λ i = λ 1. For simplicity, let i = 2. Then also ρ 2 = ρ 1, and G 1 (z) = ρ 1 1 λ 1 z 1 + ρ 1 1 λ 1 z 1 + c) Poles with multiplicity. As an example, suppose that λ 1 = λ 2 λ 3. The partial fraction expansion is ρ 1 G 1 (z) = 1 λ 1 z 1 + ρ 2 (1 λ 1 z 1 ) 2 + ρ 3 1 λ 3 z 1 The coefficients can be found by ρ 1 = 1 d λ 1 dz 1 (1 λ 1z 1 ) 2 G 1 (z) z=λ3 ρ 2 = (1 λ 1 z 1 ) 2 G 1 (z) z=λ3 ρ 3 = (1 λ 3 z 1 )G 1 (z) z=λ3 3. Calculate the inverse Z-transform for each of the terms in the partial fraction expansion. The inverse Z-transform is the sum of these (linearity). Some cases (when ROC is a region z > λ i ) a) Simple real pole ρ i 1 λ i z 1 ρ iλ n i µ[n] b) Simple complex pole (combine the conjugate complex poles in pairs) ρ i 1 λ i z 1 + ρ i 1 λ i z 1 ρ iλ n i µ[n] + ρ i (λ i ) n µ[n] 12

13 This can also be written on real form as follows. Write ρ i = ρ i e jθ i λ i = r i e jω i Then (ρ 1 λ n 1 + ρ 1(λ 1) n )µ[n] = 2 ρ 1 r n 1 cos(nω 1 + θ 1 )µ[n] c) Poles with multiplicity. Use tables, in particular Z {α n nµ[n]} = αz 1 (1 αz 1 ) 2 13

14 Algebraic Stability Test The following is equivalent to the method of section 7.9 in Mitra s textbook. Define m D m (z) = d m,k z k (NB: d m,0 = 1) k=0 D m (z) = z m D m (z 1 ) = m d m,m k z k Consider a transfer function H(z) with denominator polynomial D(z) of order M. following procedure can be used to test if all poles are inside the unit circle 1. Put D M (z) = D(z) k=0 Then the 2. Calculate the following recursively for m = N, N 1,..., 1 k m = d m,m D m 1 (z) = D m(z) k m Dm (z) 1 k 2 m 3. If for some m, k m 1. STOP. There is at least one pole not inside the unit circle. 4. If all k m < 1 all poles are inside the unit circle. Algebraic Stability Test for Complex Coefficients The procedure still works with a few modifications m D m (z) = d m,k z k k=0 D m (z) = z m D m (z 1 ) = m k=0 d m,m k z k Consider a transfer function H(z) with denominator polynomial D(z) of order M. following procedure can be used to test if all poles are inside the unit circle 1. Put D M (z) = D(z) Then the 2. Calculate the following recursively for m = N, N 1,..., 1 k m = d m,m D m 1 (z) = D m(z) k m Dm (z) 1 k m 2 3. If for some m, k m 1. STOP. There is at least one pole not inside the unit circle. 4. If all k m < 1 all poles are inside the unit circle. 14

15 Linear Phase FIR Filter H(z) = N h[n]z n : order N, length N + 1 i=0 H(z) = 0 H(1/z) = 0 Type N Relation zero at 1 zero at -1 1 even h[n] = h[n n] Possible Possible 2 odd h[n] = h[n n] Possible Y 3 even h[n] = h[n n] Y Y 4 odd h[n] = h[n n] Y Possible Type N Relation Frequency Response [ ] N/2 N [ ] N 1 even h[n] = h[n n] e jωn/2 h + 2 h 2 2 n cos(ωn) n=1 (N+1)/2 [ ] N odd h[n] = h[n n] e jωn/2 2 h n cos(ω(n )) n=1 N/2 [ ] N 3 even h[n] = h[n n] je jωn/2 2 h 2 n sin(ωn) n=1 (N+1)/2 [ ] N odd h[n] = h[n n] je jωn/2 2 h n sin(ω(n )) n=1 15

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18 Alias-free Filterbank Butterworth Any odd-order Butterworth filter with 3dB cutoff ω c = π 2 will work. Eliptic filter Any odd-order Elliptic filter with ω p + ω s = π and δ 2 s = 2δ p (1 δ p ). The method in section of the textbook can be used. FIR perfect reconstruction 1. Design Q(z) as a zero-phase, halfband filter (Nyquist) N Q(z) = q[n]z n ; n= N N must be odd 2. Let δ s = min Q(ω) 3. Find H 0 (z)h 0 (z 1 ) = F (z) as follows F (z) = Q(z) + δ s ω : F (ω) 0 { q[n] + δ s n = 0 f[n] = q[n] n 0 F (z) = δ s F (z) a) Let λ 1,..., λ 2N be the roots of F (z). b) Divide into two groups { λ 1,..., λ N } and {ˇλ 1,..., ˇλ N } so that λ i = ˇλ 1 i c) Then H 0 (z) = C N (1 λ 1 i z 1 ) 4. H 1 (z) = z N H 0 ( z 1 ) h 1 [n] = ( 1) N n h 0 [N n] 18