TLT-5200/5206 COMMUNICATION THEORY, Exercise 1, Fall 2012

Transcription

1 Problem 1. a) Derivation By definition, we can write the inverse Fourier transform of the derivative of x(t) as d d jπ X( f ) { e } df d jπ x() t X( f ) e df jπ j π fx( f ) e df Again, by definition of the Fourier inverse transform, we can identify the above expression as an inverse Fourier transform of jπ fx( f ). So based on this we get d x () t j π fx ( f ). his can be easily generalized to higher order derivatives. o be more precise, you can use the principle of mathematical induction. Base for the induction (n 1) is already derived above, then assume that n 1 d n 1 xt () ( j π f) X( f) n 1 is true and use the base and the assumption to prove that d n xt n () ( j π f ) X ( f ). n Aer that you can do the inductive conclusion that d n xt n () ( j π f ) X ( f ) n is true. 1 / 15

2 b) ime shi We can write the Fourier transform of the delayed signal xt ( ± t d ) as jπ F{ xt ( ± td)} xt ( ± td) e j π f ( t t ) x() t e ± jπ jπ e d x() t e X( f) e ± jπ d d c) Frequency shi We can write the inverse Fourier transform of the frequency shied signal X(f f c ) as 1 jπ F { X( f fc)} X( f fc) e df j π ( f ± f ) t X( f ) e ± jπ c j e X( f π ) e df ± jπ xte () From which we obtain the relation c c df F xte X f f c ± jπ c { () } ( ) / 15

3 d) ime domain multiplication By definition, we can write the inverse Fourier transform of the frequency domain convolution X(f)*V(f) as 1 jπ F { X( f) V( f)} X( f) V( f) e df jπ X( λ) V( f λ) dλe df jπ hus, X( λ) V( f λ) e dλdf j π( f λ) t X( λ) V( f + ) e dλdf jπλt jπ X( λ) e dλv( f ) e df xt () jπ x() t V( f ) e df xtvt () () F{ xtvt () ()} X( f) V( f) vt () Most of the basic Fourier transforms can be proved with simple change of variables. In addition, a very useful table of basic Fourier transforms is given in the end of Carlson 4 th edition at page / 15

4 e) Further discussion on Fourier transform duality If we have a Fourier transform x(t) X(f), then based on duality theorem there exists a transform X(t) x( f). his theorem is most useful when x(t) is real and even so X(f) is also real and even, and F{X(t)} x( f) x(f). As an example, sinc pulse is a very important function in communications theory and can be defined in time domain as z(t) Asinc(Wt) A sin (πwt) πwt We ll obtain Z(f) by applying duality to the well known transform pair of a square wave pulse v(t) v(t) BΠ t V(f) Bsinc(f) Where t is defined as follows, 1 t < t t > Rewriting z(t) as z(t) A (W)sinc(tW) W brings out the fact that z(t) V(t) with W and B A/W. Duality then says that F{z(t)} v( f) BΠ f A f Π or W W Z(f) A f Π( ) W W since the rectangle function has even symmetry. 4 / 15

5 Duality is a handy way of generating new transform pairs without the labor of integration! Also, think the duality concerning certain operations and their Fourier transforms, e.g., relation between product and convolution x(t)v(t) X(f) V(f) x(t) v(t) X(f)V(f) 5 / 15

6 Problem. he triangular pulse x(t) can be written as A(1 + t ), t < xt () A(1 t ), t <+ Now, because x(t) is symmetric wrt. origin and real-valued we can write jωt X ( f ) x() t e x()cos( t ωt) x( t)cos( ωt) A (1 t )cos( ωt) t A cos( ωt) cos( ωt) Remembering the rule for integration by parts usually stated as b ( t) v( t) u( t) v( t) a b u ' v'( t) u( t) a we can integrate also the second term and we get b a sin( ωt) tsin( ωt) sin( ωt) X ( f ) A ω ω ω sin( ω ) sin( ω ) cos( ωt) A ω ω ω 1 cos( ω ) A ω his could be the final result but we can put it in a more common form (transform tables, etc.) by using the following trigonometric identities 6 / 15

7 sin cos ( x) + cos ( x) sin ( x) 1 ( x) cos(x) Now, we can write X( f ) as ( use x ω / ) 1 cos( ω ) X( f) A ω sin ( ω ) + cos ( ω ) cos ( ω ) + sin ( ω ) A ω sin ( ω ) A ω sin ( ω ) 4A ω A A A ω sin ( ) ( ω ) π f sin ( ) ( π f) f sinc ( ) his is a sinc-function (squared) with maximum value of A at ω and zero values at ω kπ/ (or f k/), k 1,, 3, A was the triangular pulse amplitude was the wih of the pulse he following figure illustrates the absolute value of X( f ) with two different values of. 7 / 15

8 Figure 1: Absolute value of X( f ) with different values of, A1 in both cases. From the above figure we clearly see that when is increased, that is the pulse is made wider in time domain, the pulse becomes narrower in the frequency domain. his phenomenon is usually called reciprocal spreading. For the delayed version of the triangle pulse x(t-δ), we directly obtain the Fourier transform by using the fundamental Fourier theorem for time delayed signal as F{ xt ( )} X( f) e jπ f jπ f Asinc ( f) e In other words, the delay term Δ causes a constant phase shi at frequency f. 8 / 15

9 Problem 3. Again, because x(t) is real-valued and symmetric wrt. origin, we can write jωt X ( f ) x() t e x()cos( t ωt) / A (1 + cos( ωt))cos( ωt) / { cos( ω ) cos( ω )cos( ω )} A t + t t Next, we use trigonometric identity that 1 cos( x )cos( y) cos( y x) + cos( y + x) and we get / / { } { ω ω ω } X ( f ) A cos( t) + cos( t)cos( t) 1 1 A cos( ωt) + cos(( ω ω) t) + cos(( ω+ ω) t) Asin( ω ) A sin(( ω+ ω) ) A sin(( ω ω) ) + + ω ω+ ω ω ω A sin( π f) A sin( π( f + f) ) A sin( π( f f) ) + + π f 4 π( f + f) 4 π( f f) A A A sin c( f) + sin c(( f + f) ) + sin c(( f f) ) 4 4 Now, in order to the pulse to have the desired zero value at t /, it has to be fulfilled that ω π (since cos(π) -1), that is 1/( f ). So, the pulse is defined either using or f. he following figures illustrate the absolute value of X( f ) (i.e., amplitude spectrum of the pulse). 9 / 15

10 Figure : Absolute value of X( f ), also shown with dotted lines are the different components constituting to the total spectrum. 1 / 15

11 Figure 3: Absolute value of X( f ) with two different values of. 11 / 15

12 Problem 4. First we note that the impulse train n δ ( t ) δ ( t n ) is a periodic function, so let's first try to find its Fourier series (Carlson, p.4, f 1/). jπnft t) cne n δ ( where c n 1 / j nf δ ( t) e π / t he given impulse train δ (t) is actually δ(t) within the integration interval / /, so we can write c n as c n 1 1 e 1 / / δ ( t) e jπnf jπnf t where the second line follows from the integration properties of dirac-delta function (Carlson p.53). So we can write the impulse train as δ ( 1 jπnft t) e n which we can Fourier transform term by term to yield (you can always think that the spectrum of periodic signal consists of impulses weighted by the Fourier series coefficients) 1 / 15

13 1 δ () t δ( f nf) n his is the main result, the spectrum of impulse train is also an impulse train, that is, it consists of discrete frequency components. You could of course try to Fourier transform the original definition directly term-by-term to yield n δ ( t ) δ ( t n ) δ () t e n jπ fn his form, however, doesn't tell directly that the frequency domain representation is also an impulse train. We can however plot the above sum with some finite number of terms, say n 1 1 and n 1 1. his is shown in Figure 4. Clearly, as the number of terms in the sum increases, the peaks become more and more narrow. Actually, utilizing the Poisson sum formula (Carlson 3rd ed, p. 661, f λ, 1/L), we can write δ Poisson jπ fn () t e δ( f nf) n n 1 which is the same as the previous result above. ± jπnλ / L ( Poisson: e L δ ( λ ml) n m ) 13 / 15

14 Figure 4: A simple illustration of the Fourier series based representation for a periodic impulse train for a finite number of terms in the series. 14 / 15

15 his kind of a time domain impulse train n δ ( t ) δ ( t n ) is usually called ideal sampling function. his is because ideal sampling of a continuous-time signal can be considered multiplication with δ (t) where is the sampling interval. he resulting discrete-time signal consists of impulses weighted by the sample values. Now it's easy to see that the spectrum of the resulting discrete-time signal is the spectrum of the corresponding continuous-time signal with replicas at ±nf S where f S 1/. his is because multiplication in time domain becomes convolution in frequency domain and convolution with impulse train replicates the spectrum. his is illustrated in Figure 5. spectrum of x(t) -W W f spectrum of x(k) f S f S W W f S f S f Figure 5: Illustration of the spectra of continuous-time signal x(t) and its discrete-time counterpart x(k). 15 / 15