Lecture 8 ELE 301: Signals and Systems


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1 Lecture 8 ELE 3: Signals and Systems Prof. Paul Cuff Princeton University Fall 22 Cuff (Lecture 7) ELE 3: Signals and Systems Fall 22 / 37 Properties of the Fourier Transform Properties of the Fourier Transform Linearity Timeshift Time Scaling Conjugation Duality Parseval Convolution and Modulation Periodic Signals ConstantCoefficient Differential Equations Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
2 Linearity Linear combination of two signals x (t) and x 2 (t) is a signal of the form ax (t) + bx 2 (t). Linearity Theorem: The Fourier transform is linear; that is, given two signals x (t) and x 2 (t) and two complex numbers a and b, then ax (t) + bx 2 (t) ax (jω) + bx 2 (jω). This follows from linearity of integrals: (ax (t) + bx 2 (t))e j2πft dt = a x (t)e j2πft dt + b x 2 (t)e j2πft dt = ax (f ) + bx 2 (f ) Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 Finite Sums This easily extends to finite combinations. Given signals x k (t) with Fourier transforms X k (f ) and complex constants a k, k =, 2,... K, then K K a k x k (t) a k X k (f ). k= If you consider a system which has a signal x(t) as its input and the Fourier transform X (f ) as its output, the system is linear! k= Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
3 Linearity Example Find the Fourier transform of the signal This signal can be recognized as { x(t) = 2 2 t < t 2 x(t) = 2 rect ( t 2 ) + rect (t) 2 and hence from linearity we have ( ) X (f ) = 2 sinc(2f ) sinc(f ) = sinc(2f ) + 2 sinc(f ) Cuff (Lecture 7) ELE 3: Signals and Systems Fall / rect(t/2) + 2 rect(t).4.2!.2!2.5!2!.5!! sinc(ω/π) + 2 sinc(ω/(2π)).5!.5!!8!6!4! π 2π 2π ω 4π L Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
4 Scaling Theorem Stretch (Scaling) Theorem: Given a transform pair x(t) X (f ), and a realvalued nonzero constant a, x(at) ( ) f a X a Proof: Here consider only a >. (negative a left as an exercise) Change variables τ = at x(at)e j2πft j2πf τ/a dτ dt = x(τ)e a = ( ) f a X. a If a = time reversal theorem: X ( t) X ( f ) Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 Scaling Examples We have already seen that rect(t/t ) T sinc(tf ) by brute force integration. The scaling theorem provides a shortcut proof given the simpler result rect(t) sinc(f ). This is a good point to illustrate a property of transform pairs. Consider this Fourier transform pair for a small T and large T, say T = and T = 5. The resulting transform pairs are shown below to a common horizontal scale: Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
5 Compress in time  Expand in frequency.2.8 rect(t) 6 4 sinc(ω/2π) !.2!2! t!2!!5 5 π 5π 5π π ω rect(t/5) !.2!2! t! 5sinc(5ω/2π)!2 π! 5π!5 5 5π π ω Cuff (Lecture 7) ELE 3: Signals and Narrower pulse means higher bandwidth. Systems Fall / 37 Scaling Example 2 As another example, find the transform of the timereversed exponential x(t) = e at u( t). This is the exponential signal y(t) = e at u(t) with time scaled by , so the Fourier transform is X (f ) = Y ( f ) = a j2πf. Cuff (Lecture 7) ELE 3: Signals and Systems Fall 22 / 37
6 Scaling Example 3 As a final example which brings two Fourier theorems into use, find the transform of x(t) = e a t. This signal can be written as e at u(t) + e at u( t). Linearity and timereversal yield X (f ) = = = Much easier than direct integration! a + j2πf + a j2πf 2a a 2 (j2πf ) 2 2a a 2 + (2πf ) 2 Cuff (Lecture 7) ELE 3: Signals and Systems Fall 22 / 37 Complex Conjugation Theorem Complex Conjugation Theorem: If x(t) X (f ), then x (t) X ( f ) Proof: The Fourier transform of x (t) is x (t)e j2πft dt = = ( ) x(t)e j2πft dt ( x(t)e dt) ( j2πf )t = X ( f ) Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
7 Duality Theorem We discussed duality in a previous lecture. Duality Theorem: If x(t) X (f ), then X (t) x( f ). This result effectively gives us two transform pairs for every transform we find. Exercise What signal x(t) has a Fourier transform e f? Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 Shift Theorem The Shift Theorem: x(t τ) e j2πf τ X (f ) Proof: Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
8 Example: square pulse Consider a causal square pulse p(t) = for t [, T ) and otherwise. We can write this as ( ) t T 2 p(t) = rect T From shift and scaling theorems P(f ) = Te jπft sinc(tf ). Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 The Derivative Theorem The Derivative Theorem: Given a signal x(t) that is differentiable almost everywhere with Fourier transform X (f ), x (t) j2πfx (f ) Similarly, if x(t) is n times differentiable, then d n x(t) dt n (j2πf ) n X (f ) Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
9 Dual Derivative Formula There is a dual to the derivative theorem, i.e., a result interchanging the role of t and f. Multiplying a signal by t is related to differentiating the spectrum with respect to f. ( j2πt)x(t) X (f ) Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 The Integral Theorem Recall that we can represent integration by a convolution with a unit step t x(τ)dτ = (x u)(t). Using the Fourier transform of the unit step function we can solve for the Fourier transform of the integral using the convolution theorem, [ t ] F x(τ)dτ = F [x(t)] F [u(t)] ( = X (f ) 2 δ(f ) + ) j2πf = X () X (f ) δ(f ) + 2 j2πf. Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
10 Fourier Transform of the Unit Step Function How do we know the derivative of the unit step function? The unit step function does not converge under the Fourier transform. But just as we use the delta function to accommodate periodic signals, we can handle the unit step function with some sleightofhand. Use the approximation that u(t) e at u(t) for small a. Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 A symmetric construction for approximating u(t) Example: Find the Fourier transform of the signum or sign signal t > f (t) = sgn(t) = t =. t < We can approximate f (t) by the signal as a. f a (t) = e at u(t) e at u( t) Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
11 This looks like.5.5 sgn(t) e t/5 e t!.5!!.5!2!.5!! t As a, f a (t) sgn(t). The Fourier transform of f a (t) is F a (f ) = F [f a (t)] = F [ e at u(t) e at u( t) ] = F [ e at u(t) ] F [ e at u( t) ] = a + j2πf a j2πf j4πf = a 2 + (2πf ) 2 Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 Therefore, lim a(f ) a = j4πf lim a a 2 + (2πf ) 2 = j4πf (2πf ) 2 = jπf. This suggests we define the Fourier transform of sgn(t) as { 2 sgn(t) j2πf f f =. Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
12 With this, we can find the Fourier transform of the unit step, as can be seen from the plots u(t) = sgn(t) sgn(t) u(t) t t The Fourier transform of the unit step is then [ F [u(t)] = F 2 + ] 2 sgn(t) = 2 δ(f ) + ( ). 2 jπf Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 The transform pair is then u(t) 2 δ(f ) + j2πf. πδ(ω) + jω π jω ω Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
13 Parseval s Theorem (Parseval proved for Fourier series, Rayleigh for Fourier transforms. Also called Plancherel s theorem) Recall signal energy of x(t) is E x = x(t) 2 dt Interpretation: energy dissipated in a one ohm resistor if x(t) is a voltage. Can also be viewed as a measure of the size of a signal. Theorem: E x = x(t) 2 dt = X (f ) 2 df Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 Example of Parseval s Theorem Parseval s theorem provides many simple integral evaluations. For example, evaluate sinc 2 (t) dt We have seen that sinc(t) rect(f ). Parseval s theorem yields sinc 2 (t) dt = rect 2 (f ) df /2 = df /2 =. Try to evaluate this integral directly and you will appreciate Parseval s shortcut. Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
14 The Convolution Theorem Convolution in the time domain multiplication in the frequency domain This can simplify evaluating convolutions, especially when cascaded. This is how most simulation programs (e.g., Matlab) compute convolutions, using the FFT. The Convolution Theorem: Given two signals x (t) and x 2 (t) with Fourier transforms X (f ) and X 2 (f ), (x x 2 )(t) X (f )X 2 (f ) Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 Proof: The Fourier transform of (x x 2 )(t) is x (τ)x 2 (t τ) dτ e j2πft dt = x (τ) x 2 (t τ)e j2πft dt dτ. Using the shift theorem, this is = ( ) x (τ) e j2πf τ X 2 (f ) dτ = X 2 (f ) x (τ)e j2πf τ dτ = X 2 (f )X (f ). Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
15 Examples of Convolution Theorem Unit Triangle Signal (t) { t if t < otherwise. Δ(t)  t Easy to show (t) = rect(t) rect(t). Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 Since then rect(t) sinc(f ) (t) sinc 2 (f ) sinc 2 (ω/2π).5..5!!8!6!4! π 2π 2π 4π ω Transform Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
16 Multiplication Property If x (t) X (f ) and x 2 (t) X 2 (f ), x (t)x 2 (t) (X X 2 )(f ). This is the dual property of the convolution property. Note: If ω is used instead of f, then a /2π term must be included. Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 Multiplication Example  Bandpass Filter A bandpass filter can be implemented using a lowpass filter and multiplication by a complex exponential. Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
17 Modulation The Modulation Theorem: Given a signal x(t) with spectrum x(f ), then x(t)e j2πft X (f f ), x(t) cos(2πf t) 2 (X (f f ) + X (f + f )), x(t) sin(2πf t) 2j (X (f f ) X (f + f )). Modulating a signal by an exponential shifts the spectrum in the frequency domain. This is a dual to the shift theorem. It results from interchanging the roles of t and f. Modulation by a cosine causes replicas of X (f ) to be placed at plus and minus the carrier frequency. Replicas are called sidebands. Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 Amplitude Modulation (AM) Modulation of complex exponential (carrier) by signal x(t): x m (t) = x(t)e j2πft Variations: f c (t) = f (t) cos(ω t) (DSBSC) f s (t) = f (t) sin(ω t) (DSBSC) f a (t) = A[ + mf (t)] cos(ω t) (DSB, commercial AM radio) m is the modulation index Typically m and f (t) are chosen so that mf (t) < for all t Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
18 Examples of Modulation Theorem rect(t) sinc(ω/2π)!.2!2 2! 2.5!.5!!2 2! 2 t t rect(t)cos(πt)!.2!2! 2 2π π π 2π ω !.2!2! 2 2π π ( ) ω ω π 2 sinc 2π π 2π ) + 2 sinc ( ω + π 2π Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37 Periodic Signals Suppose x(t) is periodic with fundamental period T and frequency f = /T. Then the Fourier series representation is, x(t) = k= a k e j2πkft. Let s substitute in some δ functions using the sifting property: x(t) = = a k δ(f kf )e j2πft df, k= ( ) a k δ(f kf ) e j2πft df. k= This implies the Fourier transform: x(t) k= a kδ(f kf ). Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
19 ConstantCoefficient Differential Equations n k= a k d k y(t) dt k = M k= b k d k x(t) dt k. Find the Fourier Transform of the impulse response (the transfer function of the system, H(f )) in the frequency domain. Cuff (Lecture 7) ELE 3: Signals and Systems Fall / 37
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