Chapter 5 Frequency Domain Analysis of Systems

Chapter 5 Frequency Domain Analysis of Systems

CT, LTI Systems Consider the following CT LTI system: xt () ht () yt () Assumption: the impulse response h(t) is absolutely integrable, i.e., ht ( ) dt< (this has to do with system stability)

Response of a CT, LTI System to a Sinusoidal Input What s the response y(t) of this system to the input signal xt ω t θ t () = cos( + ),? We start by looking for the response y c (t) of the same system to jω t x () t e t c =

Response of a CT, LTI System to a Complex Exponential Input The output is obtained through convolution as y () t = h() t x () t = h( τ ) x ( t τ) dτ = c c c j ( t ) h( τ ω τ = ) e dτ = jωt j e h( τ ω τ = ) e dτ = = x c () t j x () t h( τ ωτ ) e d c τ

The Frequency Response of a CT, LTI System By defining it is jωτ ω τ τ H( ) = h( ) e d y () t = H( ω ) x () t = c H( ) e, t Therefore, the response of the LTI system to a complex exponential is another complex exponential with the same frequency c jω t = ω H ( ω) is the frequency response of the CT, LTI system = Fourier transform of h(t) ω

Analyzing the Output Signal y c (t) H ( ω ) Since is in general a complex quantity, we can write y () t H( ) e c = ω = jarg H( ω ) jω t = ω = H( ) e e = H ω jω t ( ) e output signal s magnitude j( ω t+ arg H( ω )) output signal s phase

Response of a CT, LTI System to a Sinusoidal Input With Euler s formulas we can express x(t) as xt () = cos( ω t+ θ ) 1 2 ( e e ) Using the previous result, the response is j( ω t+ θ) j( ω t+ θ) = + 1e jθ e jωt 1e jθ e jωt 2 2 = + 1 jθ ω jωt 1 jθ ω jωt 2 2 yt () = e H( ) e + e H( ) e

Response of a CT, LTI System to a Sinusoidal Input Cont d If h(t) is real, then H * ( ω) = H ( ω) and H( ω ) = H( ω ) e H( ω ) = H( ω ) e Thus we can write y(t) as jarg H( ω ) jarg H( ω ) 1 1 y( t) = H( ω ) e + H( ω ) e 2 2 = H( ω ) cos( ω t+ θ + arg H( ω )) j( ωt+ θ+ arg H( ω)) j( ωt+ θ+ arg H( ω))

Response of a CT, LTI System to a Sinusoidal Input Cont d Thus, the response to is which is also a sinusoid with the same frequency but with the amplitude scaled by the factor by amount xt () = Acos( ω t+ θ ) ( ) yt () = A H ( ω ) cosω t+ θ + arg H ( ω ) ω H ( ω ) arg ( ) H ω and with the phase shifted

Example: Response of a CT, LTI System to Sinusoidal Inputs Suppose that the frequency response of a CT, LTI system is defined by the following specs: H ( ω) arg H ( ω) 1.5 2 6 ω ω 1.5, ω 2, H ( ω) =, ω > 2, arg H ( ω) = 6, ω

Example: Response of a CT, LTI System to Sinusoidal Inputs Cont d If the input to the system is xt ( ) = 2cos(1t + 9 ) + 5cos(25t+ 12 ) Then the output is yt () = 2 H (1) cos(1t+ 9 + arg H (1)) + + 5 H (25) cos(25t + 12 + arg H (25)) = = 3cos(1t + 3 )

Example: Frequency Analysis of an RC Circuit Consider the RC circuit shown in figure

Example: Frequency Analysis of an RC Circuit Cont d From EEE232F, we know that: 1. The complex impedance of the capacitor is equal to 1/ jωc j t 2. If the input voltage is xc () t = e ω, then the output signal is given by 1/ jωc 1/ RC y () t = e jωt = e c R+ 1/ j C j + 1/ RC ω ω jωt

Example: Frequency Analysis of an RC Circuit Cont d Setting, it is x () t c whence we can write where ω = ω e j ω t and = and y () t c = y () t = H( ω ) x () t c H ( ) ω = j ω 1/ c RC + 1/ j RC ω 1/ RC + 1/ RC e jω t

Example: Frequency Analysis of an RC Circuit Cont d H ( ω) = ω 1/ RC + (1/ RC) 2 2 arg H ( ω) = arctan ( ωrc) 1/ RC = 1

Example: Frequency Analysis of an RC Circuit Cont d The knowledge of the frequency response H ( ω) allows us to compute the response y(t) of the system to any sinusoidal input signal since xt () = Acos( ω t+ θ ) ( ) yt () = A H ( ω ) cosω t+ θ + arg H ( ω )

Example: Frequency Analysis of an RC Circuit Cont d 1/ RC = 1 Suppose that and that xt ( ) = cos(1 t) + cos(3 t) Then, the output signal is yt ( ) = H (1) cos(1 t+ arg H (1)) + + H (3) cos(3 t + arg H (3)) = =.995cos(1t 5.71 ) +.3162cos(3t 71.56 )

Example: Frequency Analysis of an RC Circuit Cont d xt () yt ()

Example: Frequency Analysis of an RC Circuit Cont d Suppose now that xt ( ) = cos(1 t) + cos(5, t) Then, the output signal is yt ( ) = H (1) cos(1 t+ arg H (1)) + + H (5,) cos(5, t + arg H (5,)) = =.995cos(1t 5.71 ) +.2cos(5,t 88.85 )

Example: Frequency Analysis of an RC Circuit Cont d xt () yt () The RC circuit behaves as a lowpass filter, by letting lowfrequency sinusoidal signals pass with little attenuation and by significantly attenuating high-frequency sinusoidal signals

Response of a CT, LTI System to Periodic Inputs Suppose that the input to the CT, LTI system is a periodic signal x(t) having period T This signal can be represented through its Fourier series as where = x k jkω t k = xt () ce, t t + T 1 = c x x() t e jkωt dt, k k T t

Response of a CT, LTI System to Periodic Inputs Cont d By exploiting the previous results and the linearity of the system, the output of the system is yt () = H( kω ) ce k = x k jkω t x = H ( kω ) c x e j( kωt+ ar g( ck ) + arg H ( kω )) = k k = arg c y y k ck k= y y j( kωt+ arg( ck )) y jkωt, = c e = ce t k k= k

Example: Response of an RC Circuit to a Rectangular Pulse Train Consider the RC circuit with input xt () = rect( t 2 n) n

Example: Response of an RC Circuit to a Rectangular Pulse Train Cont d x( t) = rect( t 2 n) n We have found its Fourier series to be xt () = ce x jkπt, t k k with c x k 1 k = sinc 2 2

Example: Response of an RC Circuit to a Rectangular Pulse Train Cont d c x k Magnitude spectrum of input signal x(t)

Example: Response of an RC Circuit to a Rectangular Pulse Train Cont d The frequency response of the RC circuit was found to be Thus, the Fourier series of the output signal is given by H ( ω) = j ω 1/ RC + 1/ RC = ω x jkωt = y jkωt k k k= k= yt () H( k ) ce ce

Example: Response of an RC Circuit to a Rectangular Pulse Train Cont d H ( ω) ( db) 1/ RC = 1 filter more selective 1/ RC = 1 1/ RC = 1 ω

Example: Response of an RC Circuit to a Rectangular Pulse Train Cont d 1/ RC = 1 y c k y c k 1/ RC = 1 filter more selective y c k 1/ RC = 1

Example: Response of an RC Circuit to a Rectangular Pulse Train Cont d 1/ RC = 1 y() t 1/ RC = 1 y() t filter more selective 1/ RC = 1 y() t

Response of a CT, LTI System to Aperiodic Inputs Consider the following CT, LTI system xt () ht () yt () Its I/O relation is given by yt () = ht () xt () which, in the frequency domain, becomes Y( ω) = H( ω) X( ω)

Response of a CT, LTI System to Aperiodic Inputs Cont d From, the magnitude spectrum of the output signal y(t) is given by Y( ω) = H( ω) X( ω) Y( ω) = H ( ω) X( ω) and its phase spectrum is given by arg Y( ω) = arg H ( ω) + arg X( ω)

Example: Response of an RC Circuit to a Rectangular Pulse Consider the RC circuit with input xt () = rect() t

Example: Response of an RC Circuit to a Rectangular Pulse Cont d x( t) = rect( t) The Fourier transform of x(t) is ω X ( ω) = sinc 2 π

Example: Response of an RC Circuit to a Rectangular Pulse Cont d X ( ω) arg X ( ω)

Example: Response of an RC Circuit to a Rectangular Pulse Cont d 1/ RC = 1 Y ( ω) arg Y ( ω)

Example: Response of an RC Circuit to a Rectangular Pulse Cont d 1/ RC = 1 Y ( ω) arg Y ( ω)

Example: Response of an RC Circuit to a Rectangular Pulse Cont d The response of the system in the time domain can be found by computing the convolution where yt () = ht () xt () (1/ RC) t ht () = (1/ RCe ) ut () xt () = rect() t

Example: Response of an RC Circuit to a Rectangular Pulse Cont d yt () 1/ RC = 1 yt () 1/ RC = 1 filter more selective

Example: Attenuation of High- Frequency Components H ( ω) Y ( ω) = X ( ω)

Example: Attenuation of High- Frequency Components x() t y() t

The response of a CT, LTI system with frequency response to a sinusoidal signal is Filtering: if or then or Filtering Signals H ( ω) xt () = Acos( ω t+ θ ) ( ) yt () = A H ( ω ) cosω t+ θ + arg H ( ω ) ( ) H ω = H ω yt () = yt (), ( ) t

Four Basic Types of Filters lowpass H ( ω) passband highpass H ( ω) stopband stopband cutoff frequency bandpass H ( ω) bandstop H ( ω)

Filters are usually designed based on specifications on the magnitude response The phase response has to be taken into account too in order to prevent signal distortion as the signal goes through the system Phase Function arg H ( ω) If the filter has linear phase in its passband(s), then there is no distortion H ( ω)

Ideal Sampling Consider the ideal sampler:.. T xt () xn [ ] = xt ( ) = xnt ( ) t It is convenient to express the sampled signal xnt ( ) xt () pt () as where n pt () = δ ( t nt) n t= nt

Ideal Sampling Cont d Thus, the sampled waveform is = = xt pt xtδ t nt xntδ t nt () () () ( ) ( ) ( ) xt () pt () n is an impulse train whose weights (areas) are the sample values of the original signal x(t) n xt () pt () xnt ( )

Ideal Sampling Cont d Since p(t) is periodic with period T, it can be represented by its Fourier series pt () = ce jkωst, ω = k s k 2 π T sampling frequency (rad/sec) where T /2 1 c = p() t e jkωst dt, k k T T T /2 /2 1 1 = δ () te jkωst dt= T T T /2

Ideal Sampling Cont d Therefore and 1 pt () = e jkωs T k 1 1 x() t = xt () pt () = xte () ω s = xte () s T T k whose Fourier transform is 1 X ( ω) = X( ω kω ) s s T k t jk t jkω t k s

Ideal Sampling Cont d X ( ω) 1 Xs( ω) = X( ω kωs) T k

Signal Reconstruction Suppose that the signal x(t) is bandlimited with bandwidth B, i.e., Then, if the replicas of in do not overlap and can be recovered by applying an ideal lowpass filter to (interpolation filter) X( ω) =,for ω > ω 2 B, X ( ω) s 1 X ( ω) = X( ω kω ) s s T k X ( ω) X s ( ω) B

Interpolation Filter for Signal Reconstruction H ( ω) T, ω [ B, B] =, ω [ BB, ]

The impulse response h(t) of the interpolation filter is and the output y(t) of the interpolation filter is given by Interpolation Formula BT B ht () = sinc t π π yt () = ht () x() t s

But Interpolation Formula Cont d x () t = x() t p() t = x( nt) δ ( t nt) s whence Moreover, n yt () = ht () x() t = xntht ( ) ( nt) = s n BT B = xnt ( )sinc ( t nt) π π n yt () = xt ()

Shannon s Sampling Theorem A CT bandlimited signal x(t) with frequencies no higher than B can be reconstructed from its samples if the samples are taken at a rate The reconstruction of x(t) from its samples xn [ ] = x( nt ) formula xn [ ] = xnt ( ) ω = 2 π / T 2B s is provided by the interpolation BT B xt ( ) = x( nt ) sinc ( t nt ) π π n

Nyquist Rate The minimum sampling rate is called the Nyquist rate ω = 2 π / T = 2B s Question: Why do CD s adopt a sampling rate of 44.1 khz? Answer: Since the highest frequency perceived by humans is about 2 khz, 44.1 khz is slightly more than twice this upper bound

Aliasing X ( ω) 1 Xs( ω) = X( ω kωs) T k

Because of aliasing, it is not possible to reconstruct x(t) exactly by lowpass filtering the sampled signal Aliasing results in a distorted version of the original signal x(t) It can be eliminated (theoretically) by lowpass filtering x(t) before sampling it so that for Aliasing Cont d x () t = x() t p() t X ( ω ) = ω B s