Fourier Analysis of Systems Ch.5 Kamen and Heck

5.1 Fourier Analysis of Continuous- Time Systems Consider a linear time-invariant continuous-time system with impulse response h(t). y(t) = h(t) * x(t) In this chapter the system is not necessarily causal, but the impulse response is absolutely integrable—this is a stability condition.

5.1 Fourier Analysis of Continuous- Time Systems (p.2) Assume that the Fourier transform of h(t) exists and is given by H(  ). From the results of chapter 3: –Y(  )= H(  ) X(  ).(Eq. 5.4) Also we have: –|Y(  )|= |H(  )| |X(  )|. –  Y(  )=  H(  ) +  X(  ).

5.1.1 Response to a Sinusoidal Input Let x(t) = A cos(  0 t +  ), for all t. From Table 3.2, X(  ) = A  [e -j   (  +  0 ) + e j   (  -  0 ) ] Now H(  )  (  + c) = H(-c)  (  + c) And so we have: Y(  )= H(  ) X(  ) = A  H(  )[e -j   (  +  0 ) + e j   (  -  0 ) ] = A  [H(-  0 )e -j   (  +  0 ) +H(  0 )e j   (  0 -  0 )]

5.1.1 Response to a Sinusoidal Input (p.2) Also, h(t) is real valued and so |H(-  0 )| = |H(  0 )|  H(-  0 ) = -  H(  0 ) (Eq. 5.9) This then gives the Fourier transform of the output to be Y(  ) = A  |H(  0 )| [e –j(  + H(  0 ) )  (  +  0 ) +e j(  +  H(  0) )  (  0 -  0 )](Eq. 5.10) From the Table of inverse transforms: y(t) = A |H(  0 )| cos(  0 t +  +  H(  0 ) )

Example 5.1 Response to Sinusoidal Inputs Let the frequency response function be given by a magnitude function and phase function: –|H(  )| = 1.5 for 0    20 and 0 for  >20 –  H(  ) = - 60  for all . If the input is: –x(t) = 2 cos(10t + 90  ) + 5cos(25t  ) for all t. Then the output is: –y(t) = 3 cos(10t + 30  ) for all t.

Example 5.2 Frequency Analysis of an RC Circuit See Figure 5.1. Figure 5.2 and Figure 5.3 show results.

Example 5.3 Mass-Spring – Damper System Section 1.4, Figure M y’’(t) + D y’(t) + K y(t) = x(t) (5.23) M is the mass. D is the damping constant. K is the stiffness constant. x(t) is the force applied to the mass. y(t) is the displacement of the mass relative to the equilibrium position.

Example 5.3 (cont.) Take the Fourier transform of both sides of the differential equation: –M(j  ) 2 Y(  ) + D(j  )Y(  )+ KY(  ) = X(  ) –Y(  )(M(j  ) 2 + D(j  )+ K) = X(  ) –Y(  )= X(  ) / [(M(j  ) 2 + D(j  )+ K)] –Y(  )= H(  ) X(  ) –where H(  )= 1/(M(j  ) 2 + D(j  )+ K)

5.2 Response to Periodic and Nonperiodic Inputs Periodic Inputs –x(t) = a 0 +  k=1,  A k cos(k  0 t +  k ) for all t –y(t) =H(0) a 0 +  k=1,  |H(k  0 )|A k cos(k  0 t +  k +  H(k  0 )) for all t (Eq. 5.24) –Example 5.4 Response to a Rectangular Pulse Train Nonperiodic Inputs –y(t) = inverse Fourier Transform of H(  )X(  ). –Example 5.5 Response of RC Circuit to a pulse.

5.3 Analysis of Ideal Filters Figure 5.12 illustrates the magnitudes of 4 ideal filters. –Lowpass –Highpass –Bandpass –Bandstop More complicated ideal filters can be obtained by cascading the above: –Figure 5.13—Ideal Comb Filter

5.3.1 Phase Functions To avoid phase distortion, the ideal filter should have linear phase over the passband of the filter. –  H(  ) = -  t d for all  in the passband. –t d is a fixed positive number that represents a time delay through the filter.

5.3.2 Ideal Linear-Phase Lowpass Filter H(  ) = e –j  t d, -B  B, and 0 elsewhere. The input response can be found by finding the inverse of the frequency response: Rewrite frequency response: H(  ) = p 2B e –j  t d From Table 3.2: (  /2  ) sinc(  t/2  )  p  (  ) Let  = 2B, (2B/2  ) sinc(2Bt/2  )  p 2B (  ) Apply time shift: –(2B/2  ) sinc(2B(t-t d )/2  )  p 2B e –j  t d Hence: h(t) = (B/  ) sinc(B(t-t d )/  ) for all t. For other ideal filters, the analysis is similar.

5.4 Sampling Let p(t) =  n=- ,   (t-nT) Eq Then the sampled waveform is –x(t)p(t) =  n=- ,  x(t)  (t-nT)=  n=- ,  x(nT)  (t-nT) Determine the Fourier transform of the sampled signal: –Reconsider the pulse waveform: (page 243) p(t) =  k=- ,  c k e jk  s t, where  s = 2  /T is the sampling frequency. Then the Fourier coefficients are c k =1/T

5.4 Sampling (p.2) Now, x s (t) =x(t)p(t) =  k=- ,  x(t)(1/T)e jk  s t Use the property of multiplication by a complex exponential: –x(t)e j  0 t  X(  -  0 ) Thus, the transform of the sampled signal: –Xs(  ) =  k=- ,  (1/T)X(  - k  S ) See Figure 5.17 for the case where the spectral replicas do not overlap.

5.4.1 Signal Reconstruction Suppose that the orignal signal, x(t), is bandlimited: |H(  )|= 0 for  > B. If the sampling frequency,  s  2B, then the replicas in Xs(  ) will not overlap. Now consider and ideal low-pass filter as shown in Figure 5.18 (page 245). If the sampled signal is passed through the ideal-low pass filter, the only component passed is X(  ).

Sampling Theorem A signal with bandwidth B can be reconstructed completely and exactly from the sampled signal x s (t) = x(t)p(t) by lowpass filtering with cutoff frequency B is the sampling frequency if the sampling frequency  s is chosen to be greater than or equal to 2B. The minimum sampling frequency is called the Nyquist sampling frequency.

5.4.2 Interpolation Formula Pages 246 and 247 goes through the formal discussion of passing the sampled signal through an ideal lowpass filter. The final equation, (5.57), is called the interpolation formula.

5.4.3 Aliasing If the replicas overlap, reconstruction results with high frequency components being transposed to lower frequency components—this is called aliasing. Figure 5.20 and 5.21 illustrate the concept. Example 5.6 and 5.7 discuss sampling of speech. Note: in the real-world an ideal aliasing filter cannot be built, so there will always be some distortion.

5.5 Fourier Analysis of Discrete- Time Systems Convolution: y[n] = h[n] * x[n] Frequency Response: H(Ω) (DTFT of h[n] Y(Ω) = H(Ω) X(Ω)

5.5.1 Response to a Sinusoidal Input Let x[n] = A cos(Ω o n + θ), n=0, ±1,±2 … Take DTFT of x[n]. Multiply by frequency response. Take inverse DTFT (see page 250) y[n] = A ǀ H(Ω o ) ǀ cos(Ω o n + θ + H(Ω o )) n=0, ±1,±2 …

Example 5.8 Response to a Sinusoidal Input Let H(Ω) = 1 + e -jΩ Let x[n] = x1[n] + x2[n] = sin(π/2 n) Then H(0) = = 2 And H(π/2) = 1 + exp(-jπ/2) = –j So, y[n] = 2 + (2) (√2)sin(π/2 n - π/4).

Example 5.9 Moving Average Filter Let y[n] = (1/N){x[n] + x[n-1]+…+x[n-(N-1)]} Take DTFT and use time-shift property. Y(Ω) = (1/N){X(Ω) + X(Ω)e -jΩ +…} Y(Ω) = (1/N){1 + e -jΩ +…+e -j(N-1)Ω } X(Ω) So H(Ω) = (1/N){1 + e -jΩ +…+e -j(N-1)Ω } And H(Ω) = {sin(NΩ/2)/N sin(Ω/2)}e -j(N-1)Ω/2 Figure 5.22 shows frequency response for N=2.

5.6 Application to Lowpass Digital Filtering Analysis of Ideal Lowpass Digital Filter See Figure 5.23 (phase =0) Let x[n] = A cos(Ω o n), n=0,±1,±2,… Then y[n] = A cos(Ω o n), n=0,±1,±2,… for 0≤ Ω o ≤B and 0 elsewhere.

5.6.2 Digital-Filter Realization of Ideal Analog Lowpass Filter Let x(t) = A cos(ω o t), -∞≤ t ≤∞ Sample the signal: x[n] = A cos(Ω o n) where Ω o = ω o T and t = nT. For the ideal filter we must have Ω o <π or ω o <π/T. An analog signal can then be generated from the sampled output. For this filter, h[n] = (B/π) sinc(Bn/π)