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(0t + ), 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()

= AH()[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 beY() = 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(0t + + 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 + 120) 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 1.25.• 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)2Y() + 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) = a0 + k=1, Ak cos(k0t +k) for all t

– y(t) =H(0) a0 + k=1, |H(k 0)|Ak cos(k0t +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() = -td for all in the passband.

– td is a fixed positive number that represents a time delay through the filter.

5.3.2 Ideal Linear-Phase Lowpass Filter

• H() = e –jtd, -BB, and 0 elsewhere.• The input response can be found by finding the

inverse of the frequency response:

• Rewrite frequency response: H() = p2Be –jtd

• From Table 3.2: (/2) sinc(t/2)p()

• Let = 2B, (2B/2) sinc(2Bt/2)p2B()

• Apply time shift:– (2B/2) sinc(2B(t-td)/2) p2Be –jtd

• Hence: h(t) = (B/) sinc(B(t-td)/) for all t.

• For other ideal filters, the analysis is similar.

5.4 Sampling

• Let p(t) = n=-, (t-nT) Eq. 5.47

• 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=-,ckejkst , where s = 2/T is the sampling frequency.

• Then the Fourier coefficients are ck=1/T

5.4 Sampling (p.2)

• Now, xs(t) =x(t)p(t) = k=-,x(t)(1/T)ejkst

• Use the property of multiplication by a complex exponential: – x(t)ej0t X( - 0)

• Thus, the transform of the sampled signal:– Xs() =k=-,(1/T)X( - kS)

• 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 xs(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(Ωo n + θ), n=0, ±1,±2 …

• Take DTFT of x[n].

• Multiply by frequency response.

• Take inverse DTFT (see page 250)

• y[n] = A ǀH(Ω o)ǀ cos(Ωo n + θ + H(Ωo)) 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] = 2 + 2 sin(π/2 n)

• Then H(0) = 1 + 1 = 2

• And H(π/2) = 1 + exp(-jπ/2) = 1 + 0 –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

• 5.6.1 Analysis of Ideal Lowpass Digital Filter

• See Figure 5.23 (phase =0)

• Let x[n] = A cos(Ωon), n=0,±1,±2,…

• Then y[n] = A cos(Ωon), n=0,±1,±2,… for 0≤ Ωo≤B and 0 elsewhere.

2( ) ( 2 )B

k

H P k

5.6.2 Digital-Filter Realization of Ideal Analog Lowpass Filter

• Let x(t) = A cos(ωot), -∞≤ t ≤∞

• Sample the signal: x[n] = A cos(Ωon) where Ωo = ωoT and t = nT.

• For the ideal filter we must have Ωo <π or ωo<π/T.

• An analog signal can then be generated from the sampled output.

• For this filter, h[n] = (B/π) sinc(Bn/π)