Ver 3808 E1.10 Fourier Series and Transforms (2014) E1.10 Fourier Series and Transforms. Problem Sheet 1 (Lecture 1)

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1 Ver 88 E. Fourier Series and Transforms 4 Key: [A] easy... [E]hard Questions from RBH textbook: 4., 4.8. E. Fourier Series and Transforms Problem Sheet Lecture. [B] Using the geometric progression formula, evaluate 5 r r.. [B] Determine expressions not involving a summation for a r xr, b r x r, c R r xr y r, d R r r.. [B] In the expression 5 r r, make the substitution r m + and then evaluate the resultant expression. 4. [C] Determine a simplified expression not involving a summation for N r N ejωr for N. 5. [C] Determine a simplified expression for R r ejπrr for R. Ensure your answer is correct even when R. 6. [C] Determine the value of N n M m xm n. 7. [C] If xt sin t, determine a xt, b xt and c x t where denotes the timeaverage. 8. [C] The first two normalized Legendre polynomials are P t and P t t. a Show that P t [,] P t [,] and P tp t [,] where [,] denotes the average over the interval < t <. b If P t at + bt + c, find the coefficients a, b and c such that P tp t [,] P tp t [,] and P t [,] Problem Sheet Page of

2 Ver 98 E. Fourier Series and Transforms 4 E. Fourier Series and Transforms Problem Sheet - Solutions. 5 r r In the expression 5, the 5 is the number of terms in the sum and the is the first term when r.. a Each term is multiplied by a factor of x, so the standard formula gives x x x where x x since there are terms. b Each term is multiplied by a factor x and, treating x, the first term equals, so the sum is x x. c Each term is multiplied by xy and the first term is y so the sum is y xyr+ xy. d Each term is multiplied by and the first term is so the sum is R+ + R R even R odd.. The substitution r m + m r. So, making the substitution in both the limits and summand gives r m+ m 5. r So the answer is 6 as in question. m 4. Each term is multiplied by e jω and the first of the N + terms is e jωn so the sum is m jωn ejωn+ e e jω e jωn e jωn+ e jω. A very common trick when an expression includes the sum or difference of two exponentials is to take out a factor whose exponent is the average of the two original exponents; in this case the average exponent is j.5ω in the denominator and also in the numerator since jωn+jωn+ j.5ω. This gives e j.5ω e jωn+.5 e jωn+.5 e j.5ω e j.5ω e j.5ω e jωn+.5 e jωn+.5 e j.5ω e j.5ω j sin N +.5ω j sin.5ω 5. Each term is multiplied by e jπr and the first term is e so the sum formula gives e jπrr ejπ. e jπr e jπr e jπr sin N +.5ω sin.5ω However, when R, the denominator is zero so the formula is invalid; in this case there is only one term in the summation and it equals e jπ. So the answer is for all values of R except R when the answer is. We can write this compactly as R δ[r ] R > where the function δ[n] is the Kroneker Delta function and equals if and only if its integer argument equals zero. 6. The summand in this question is separable because it can be expressed as the product of two factors that depend on m and n respectively. So we can write N M N x m n n m n x n M m x m x N+ x x xm x x x N+ x M x. Solution Sheet Page of

3 Ver 98 E. Fourier Series and Transforms 4 7. athe period is π, so we calculate the average by integrating over one period and dividing by the period: xt π π sin t dt π [ cos t]π. b The period is now π, so we calculate the average as: xt π π sin t dt π π sin t dt π [ cos t]π π. c The period is still π: x t π ˆ π sin t dt π ˆ π cos t dt π An easier way of getting this answer is to write ˆ π sin t cos t. cos t dt [ t ] π π sin t π π. 8. a and Finally P t [,] P t [,] P tp t [,] ˆ ˆ dt [t] t dt ˆ t dt b The analysis is slightly easier if you do it in the right order. P tp t [,] ˆ [ t ]. 4 [ t ]. [ at at + bt 4 + ct dt 4 + bt + ct ] b so b. Now ˆ P tp t [,] at + c dt [ ] at + ct a + c so a c. Finally P t [,] c ˆ 9t 4 6t + dt c [ 9t 5 5 t + t ] c c from which c ± 5. So the polynomial is P t 5 t. Solution Sheet Page of

4 Ver 54 E. Fourier Series and Transforms 4 Key: [A] easy... [E]hard E. Fourier Series and Transforms Problem Sheet Lectures, Questions from RBH textbook:.,.,.,.4,.5,.8,.9,.,.,.,.,.4,.5,.7,.,.,.,.6.. [B] Give the fundamental period of a cos πt, b cos πt +. cos 5πt, c cos πt + cos t.. [C] A sufficient condition for a periodic function, ut, to have a Fourier series is that it satisfies the Dirichlet conditions on page -5 of the notes. Determine which of the following functions satisfies these conditions. The notation x mod n means the remainder when x is divided by n. a sin t, b sin t, c sin t, d +t, e t mod.. [B] Determine the fundamental frequency and the Fourier Series coefficients for ut + cos6πt+ sin 4πt. 4. [B] The phasor + 4i represents the waveform cos ωt 4 sin ωt. Give a the Fourier coefficients and b the complex Fourier coefficients for this waveform. 5. [C] Determine the fundamental frequency and the Fourier Series coefficients for ut cos 4 πt. 6. [C] a Determine the Fourier coefficients, a n, b n } for the waveform, ut, with period T defined by ut t for t <. b Determine the complex Fourier coefficients, U n, for the same waveform. c Determine the complex Fourier coefficients for vt ut. d Determine the complex Fourier coefficients for wt vt [B] If ut has period T F and Fourier coefficients a : [5,, ] and b with all other coefficients zero. a Give an expression for ut, b Determine the complex Fourier coefficients, U n. 8. [B] Each of the following waveforms has period T and equals the expression given for t <. In each case say whether the complex Fourier coefficients will be i real-valued, ii purely imaginary or iii neither. a t b t c t+t d t + e t + f t sin t g t cos t h t sin t. 9. [C] Each of the following waveforms has period T and equals the expression given for t <. In each case say whether or not all the even-numbered Fourier coefficients will equal zero. a sin πt b t + t < t + t < c t t t t d t t e t t.. [C] ut has period T 4 and is defined by ut t < t < 4. a Find the complex Fourier coefficients, U n expressing them in polar form: r e iθ. Identify which of the coefficients are equal to zero. b Find the complex Fourier coefficients of vt ut +.5 and explain why they are necessarily real-valued. Explain the relation between the magnitudes V n and U n. c Find the complex Fourier coefficients of wt vt + vt. Identify which of the coefficients are non-zero and explain how your answer relates to the symmetries of wt. Problem Sheet Page of

5 Ver 5546 E. Fourier Series and Transforms 5 E. Fourier Series and Transforms Problem Sheet - Solutions. a The fundamental frequency is π π 5 Hz and the period is π π ms. b For a mixture of cosine waves, the fundamental period is the lowest common multiple LCM of the periods of the constituent waves. In this case the frequencies of the two waves are 5 Hz and 65 Hz with periods ms and.6 ms respectively. The LCM of and.6 is 8 ms corresponding to a frequency of 5 Hz. Alternatively, you can obtain the same answer by finding the highest common factor HCF of the two frequencies. Notice that the addition of even a tiny amplitude at 65 Hz has quadrupled the fundamental period. c The periods of the two constituent waves are here ms and ms. Since the second of these is irrational, there is no LCM and the resultant waveform is not periodic at all or equivalently its period is.. a Yes. b No. We have T π, so T sin t π dt sin t dt [ln tan.5t]π ln tan.5π tan c Yes for a similar reason to the previous part since T τ dτ [ τ] T T <. d No since it is not periodic. e Yes; this function is a triangle wave with period. π ln.. The fundamental frequency is the highest common factor of the constituent frequencies, i.e. π rad/s khz. So, setting F, we have ut + cos πf t + sin πf t. So the Fourier coefficients are a, b, a with all other coefficients zero. 4. a The non-zero Fourier coefficients are a and b 4. b The non-zero complex Fourier coefficients are U i and U + i. We see that U is exactly half the value of the phasor and that U is the complex conjugate of U. 5. We need to express cos 4 θ in terms of components of the form cos nθ. We can do this by writing cos 4 θ e iθ + e iθ 4 6 e i4θ + 4e iθ e iθ + e i4θ 6 cos 4θ + 8 cos θ From this we find that the fundamental frequency is actually 4π rad/s khz and the non-zero Fourier coefficients are therefore a 4, a and a a From the formulae on slide - of the notes and observing that F T, a n T b n ˆ ˆ ut cos πnf t dt t cos πnt dt π n [πnt sin πnt + cos πnt] t ˆ t sin πnt dt π n [ πnt cos πnt + sin πnt] t 6 6 n cos πn πn πn Note that the b n expression applies only for n. Notice also that the a n are all zero because ut is a real-valued odd function and that the coefficient magnitudes are n which is a characteristic of waveforms that include a discontinuity. Solution Sheet Page of

6 Ver 5546 E. Fourier Series and Transforms 5 b We have U a and, for n, U ±n a n ib n ±i n from which U n i n πn. Again, we note that U n is purely imaginary because ut is a real-valued odd function. c We can calculate the V n directly by integrating over an interval that does not include a discontinuity. V for n : V n ˆ ˆ ˆ vtdt vte iπnt dt π n ˆ t dt t e iπnt dt [ + iπn t e iπnt ] π n t iπn π n i πn An alternative way to calculate V n is to use the time-shifting formula: vt ut V n U n e iπnf U n e inπ n U n. Notice that time-shifting a waveform changes the phases of the V n but not the magnitudes. d The complex Fourier transform of xt 4 is iust X 4 with all other coefficients zero. So, since the Fourier transform is linear, if wt vt + xt we must have W n V n + X n which means that W 4 and, for n, W n 6i πn. 7. a ut.5 + cos πf t + sin πf t + cos 4πF t. b U ±n a n ib n from which U : [.5, +.5i,.5,.5i,.5]. Notice that, since ut is real, U n is the complex conjugate of U +n. 8. The Fourier transform of a real-valued signal is purely real or purely imaginary if it is even or odd respectively. So we have the following: a real b imaginary c neither d real e neither f real g imaginary g imaginary. 9. All the even-numbered Fourier coefficients of a waveform are zero if it is anti-periodic which, in this case with T, means that ut ut for t < ; note that you only need to prove this relationship for half a period since the periodicity relationship, ut ut + T, then means that it applies for the other half. So we have the following Yes means it is anti-periodic: a Yes: sin πt sin πt π and of course the Fourier transform has only a single component with b or, equivalently, X ± i. b Yes: for t <, ut t and ut t + t ut c No: for t <, ut t but ut t + t ut. Note however that vt ut.5 is anti-periodic since, for t <, vt ut.5 t +.5 t.5 ut.5 vt. Thus the only non-zero even harmonic of ut is U.5. d Yes: for t <, ut t t and ut t + t t t ut e No: for t <, ut t t but ut t t t t t + t ut. Solution Sheet Page of

7 Ver 5546 E. Fourier Series and Transforms 5. a We note that the fundamental frequency is F 4. U n 4 ˆ e i.5πnt dt i e i.5πn 4.5πn ie i.5πn πn ie i.5πn e i.5πn e i.5πn i sin.5πn πn sin.5πn e i.5πn πn We know that sin θ whenever θ is a multiple of π, so U n whenever n is a non-zero multiple of 4. sin.5πn b By the time-shift formula V n U n e iπnf.5 U n e i.5πn πn. Since vt is real and symmetric, V n will also be real and symmetric. The time-shifting affects only the phase and so V n U n. c By linearity and the time-shift formula, W n V n + e iπnf V n + e iπn V n + n. The quantity + n equals for even values of n and for odd values of n. Since in addition, V n when n is a non-zero multiple of 4, W n is only non-zero for n and for odd multiplies of. In fact, the period of wt is rather than 4 which explains why W n for all odd values of n. In addition, when considered with a period of, wt W is anti-periodic and so all its even Fourier coefficients will be zero. Solution Sheet Page of

8 Ver 54 E. Fourier Series and Transforms 4 Key: [A] easy... [E]hard E. Fourier Series and Transforms Problem Sheet Lectures 4, 5 Questions from RBH textbook:.9,.,.5.. [C] a Determine the fundamental frequency, the Fourier coefficients and the complex Fourier coefficients of ut cos t. b Determine the power, P u u t where denotes the time average. Hint: cos 4 t 8 cos 4t + cos t + 8. c Show that Parseval s theorem applies: P u n U n 4 a + n a n + b n.. [C] The even function ut with period T is defined in the region t by ut a t a t > a where < a <. a Determine the complex Fourier coefficients, U n. b Explain why U does not depend on a. c Show that sin anπ n anπ a.. [C] Determine the fundamental frequency and the complex Fourier Series coefficients of xt cos 8πt cos πt in two ways: a by expanding our the product using trigonometrical formulae and b by convolving the Fourier coefficients of the two factors. 4. [C] a Give the complex Fourier coefficients, U n, if ut cos t. b Show, by using the convolution theorem, that vt u t cos t +. c Show, by using the convolution theorem again that wt v t u 4 t 8 cos 4t + cos t + 8. t < π 5. [C] Suppose ut sin t and vt both with period T π. π t < π a Determine the complex Fourier coefficients U n and V n. b If wt utvt, determine W n U n V n by convolving U n and V n. 6. [B] The waveform ut has period T and equals ut 4t for t <. If u N t N n N U ne iπnt estimate, for large N, the minimum value and maximum value of u N t and also the value of u N. 7. [B] The waveform ut has period T. Estimate how rapidly U n will decrease with n when ut in the range t < is given by a t, b t, c t t, d t t, e t t + t + 8. [C] The waveform ut has period T u and satisfies ut exp t for t <. The waveform vt has period T v and satisfies vt exp t for t <. a Find expressions for the complex Fourier coefficients U n and V n. b Calculate the average powers u t and v t and also those of u t and vt where u N t is the waveform formed by summing harmonics N to +N. c Determine the average error powers ut u t and vt v t. Problem Sheet Page of

9 Ver 5594 E. Fourier Series and Transforms 5 E. Fourier Series and Transforms Problem Sheet - Solutions. a We have ut cos t + cos t. So the fundamental period is T π and the fundamental frequency is F T π. The Fourier coefficients are a and a, so the complex Fourier coefficients are U, U U 4. b P u π π cos4 t dt π [t + 8 sin t + sin 4t]π π π c n U n Also 4 a + n a n + b n Note that the formula for Parseval s theorem is much more elegant and memorable when using complex Fourier coefficients.. a We have U n T ˆ T T ˆ a a ute iπnf t dt a e iπnt dt i [ e iπnt ] a anπ t a i e iπna e iπna anπ sin anπ anπ Note that U n is real-valued and even as expected since ut is real-valued and even. b From the formula U n but this is not defined so we either determine U directly sin anπ anπ T T from the original integral as U sin anπ T ut dt or else as a limit: U lim n anπ. We can sin anπ aπ cos anπ find this limit using L Hôpital s rule: lim n anπ aπ n or, equivalently, by using sin anπ the small angle approximation, sin x x, which is exact for x and gives U lim n anπ anπ anπ. It is always true that U ut so since the average value of ut is for all values of a, it follows that U will not depend on a. c We can calculate ut ˆ T T w ˆ w w T u t dt w dt So, by Parseval s theorem, we know that U n n sin wnπ wnπ n ut w. a Expanding the product gives xt 6 cos πt + 4 cos 8πt cos πt 6 cos πt + cos πt + cos 8πt. The fundamental frequency is the HCF of the frequencies of these three components or, equivalently, of the original two components and equals Hz or 4π rad/s. The three frequency components ar therefore at 5, and 7 times the fundamental frequency giving the coefficient set: X 7:+7 [,,,,,,,,,,,,,, ]. Note that since xt is even, the coefficients are Solution Sheet Page of

10 Ver 5594 E. Fourier Series and Transforms 5 are symmetrical around X which is underlined. b We can write xt utvt where ut cos 8πt and vt cos πt. Using the fundamental frequency of the output i.e. Hz, the coefficients of ut and vt are U : [,, 6,, ] and V 5:5 [.5,,,,,,,,,,.5]. To convolve these, we replace each non-zero entry in V 5:5 with a complete copy of U : scaled by the corresponding entry of V 5:5. This gives the same coefficients as in the previous part. 4. a The only non-zero coefficients are U ±.5. b Convolving U n with itself gives V ±.5 and V Thinverse Fourier transform gives vt cos t + as required. c Convolving V n with itself gives W ±4.5.65, W ± and W Taking the inverse Fourier transform gives the required answer. 5. a U i and U i. For V n we write V π for n : V n π i nπ i nπ i i nπ ˆ π ˆ π e it dt e int dt [ e int ] π e inπ nπ n n odd n even, n n Note that, except for its DC component of V, vt is a real-valued, odd, anti-periodic function and therefore has purely imaginary coefficients with all even coefficients except V equal to zero. b From the notes slide 4-5 the convolution is defined by W n U n V n V n U n m V n mu m. Since U m except for m ±, the infinite sum actually only has two non-zero terms and W n U V n +U V n+ i V n+ V n. If n is even, then n+ and n are both odd so, using the formula for V n given above, W n i V n+ V n i i n+π i n π π n+ n π n n π. If n is odd then n + and n are both even and V n+ and V n are both zero unless n + or n equals zero, i.e. unless n ±. So we have W i V i 4 and W i V i 4. We can combine all these results to give n odd, n ± i W n 4n n ± n π n even 6. We have u u but u+ so there is a discontinuity at t. Therefore u N +. Notice that the actual value defined for u has no affect on this answer. Due to Gibbs phenomenon, u N t will undershoot and overshoot the discontinuity by about 9% of the discontinuity height: 4. So So the maximum value of u N t will be.6 and the minimum value will be a u but u so the waveform has a discontinuity and the coefficients, U n, will decrease n. b u but u so the waveform again has a discontinuity and the coefficients, U n, will decrease n. c u u but u u so coefficients, U n, will decrease n. Solution Sheet Page of

11 Ver 5594 E. Fourier Series and Transforms 5 d The first non-equal derivative is u u so coefficients, U n, will decrease n. e u u and u u. The first non-equal derivative is 6 u u 6 so coefficients, U n, will decrease n. 8. a U n T u et e iπnfut dt [ e iπnt dt ] iπn e iπnt t iπn e iπn iπn e e iπn iπn e e iπn. Note that we use the fact that e iπn for any integer n. V n T v e t e iπnfvt dt e t e iπnt dt + iπn e iπn + iπn e iπn iπn e n + iπn e n et e iπnt dt n e +iπn + iπn n e +π n n e +π n. We see that this is real-symmetric because vt is real-symmetric and that it decays n because vt is continuous but has gradient discontinuities at t and t. b u t T u et dt [ et dt ] e t t e.945. v t u t e since reflecting a waveform in time does not affect its power. u t U n U + U + U v t V n V + V + V u t ut v t ut We see that, for the same number of harmonics, v t fits the exponential much better than u t over the range t < and that it includes much more of the energy of ut. c We can use Parseval s theorem to calculate the power of the error, ut u t. We know that ut + U ne iπnt and that u t + U ne iπnt, so it follows that ut u t n > U ne iπnt. Applying Parseval s theorem to these threee expressions gives u t + U n, u t + U n and ut u t n > U n. By subtracting the first two of these equations, we can see that u t u t ut u t and so, from part b, u t u t Likewise ut u t vt v t Note that, for arbitrary functions xt and yt having the same period, the relationship xt yt x t y t is only true if xtyt or, equivalently, if they have non-overlapping Fourier series i.e. X n and Y n are never both non-zero for any n. Solution Sheet Page of

12 Ver 5489 E. Fourier Series and Transforms 4 Key: [A] easy... [E]hard E. Fourier Series and Transforms Problem Sheet 4 Lectures 6, 7, 8 Fourier Transform: Xf xte iπft dt Inverse Transform: xt Xfeiπft df Questions from RBH textbook:.,.,.,.5,.7,.9,.9,... [B] Evaluate δt t e t dt.. [B] a Evaluate δt 6t dt. b Now make the substitution t τ for the integration variable and show that the integral remains unchanged.. [B] Express x δ8 x in the form aδx b 4. [C] a If vt e t, show that its Fourier transform is V f +4π f. b Using the time shifting and scaling formulae from slides 6-9 and 6- and without doing any additional integrations, determine the Fourier transforms of i v t e at, ii v t e t b, iii v t +t. 5. [D] Determine the Fourier transform, Xf, when xt t e t. 6. [C] If xt δt determine the Fourier transform, Xf. Hence, by considering the inverse transform, show that eiαft df π α δt where α is a real constant. 7. [B] Determine the Fourier transform, Xf, when xt is a DC voltage: xt. 8. [B] Determine the Fourier transform, Xf, when xt cos πt + 8 sin 4πt. 9. [C] If vt is a periodic signal with frequency F for which vt δt for F t < F, determine the coefficients, V n, of its complex Fourier series. Hence find the Fourier transform, Xf, of the impulse train given by xt n δt n F.. [C] If the Fourier transform of xt is Xf cos f, determine xt in two ways: a using the duality relation: vt Ut V f u f and b by directly evaluating the inverse transform integral and using the result of question 6.. [B] If xt or rectt.. [B] If xt t.5 t >.5 e at t t < sin πf show that Xf πf. This function is often called a top-hat function show that Xf iπf+a for a >.. [C] An electronic circuit, whose input and output signals are xt and yt respectively, has a frequency response given by Y X iω iω+. a If xt cos t, use phasors to find an expression for yt. Give expressions for the Fourier transforms Xf and Y f. e 5t t b If xt give an expression for Y f you may use without proof the result of t < question. Show that Y f may be written as constants c and d. Hence give an expression for yt. 4. [C] The triangle function is given by yt c iπf+5 + d iπf+ and find the values of the t t. Show that yt may be obtained by t > convolving xt with itself where xt rectt as defined in question, i.e. yt xt xt xτxt τdτ. Hence use the convolution theorem and the result of question to give the Fourier transform Y f. Problem Sheet 4 Page of

13 Ver 5489 E. Fourier Series and Transforms 4 5. [B] An energy signal has finite energy: E x xt dt <. A power signal has infinite energy but finite power: xt B lim A,B B A A xt dt <. Say whether each of the following functions of time, t, is i an energy signal, ii a power signal or iii neither: a cos ωt, b, c t, d t, e e t, f e t, g e t, h +t, i cos t, i + t, k. t 6. [C] Suppose the Fourier transform of xt is Xf + i δf + 4 δf 4. Give expressions for the alternative versions of the Fourier transform: a Xω +πf xte iωt dt and b Xω π xte iωt dt. State the general formulae for the inverse transform integrals that give xt in terms of Xω and Xω. Problem Sheet 4 Page of

14 Ver 54 E. Fourier Series and Transforms 4 E. Fourier Series and Transforms Problem Sheet 4 - Solutions. δt t e t dt [ t e t] t e a δt 6t dt [ t ] t6 6 b Substituting t τ gives δτ 69τ dτ 7 δ τ τ dτ 7 δτ τ dτ 9 [ τ ] 6. We here use the relation that c δcx δx. τ. x δ8 x x δ x 4 x δ x 4 x δ x 4 6δ x a V f e t e iπft dt et e iπft dt + e t e iπft dt [ ] iπf e iπft t + iπf [ e iπft ] t iπf iπf +4π f. Notice that in the first step we split the integral up into the two ranges of t for which the quantity t is equal to t and +t respectively; this is necessary for any integral involving absolute values. Also notice that e a+bit is zero at t + if a < and zero at t if a >. b If v t vat then V f a V f a a a +4π f. If v t vt b then V f e iπfb V f e iπfb +4π f. If wt V t +4π t then W f v f e f. However we want v t.5w t π so V f.5 π W πf πe πf. 5. Xf ˆ ˆ ˆ t e t e iπft dt t e t e iπft dt + t e iπft dt + ˆ ˆ t e t e iπft dt t e iπft dt [ iπf e t iπft iπf t + iπf ] t ] [ + iπf e t iπft iπf t + iπf iπf iπf π f + 4π f t 6. Xf xte iπft dt δte iπft dt [ e iπft]. Note that this is the same for all t values of f and is called a flat or white spectrum. The inverse transform is δt ˆ Xfe iπft df ˆ e iπft df. If we now substitute τ π α t, we obtain eiαfτ df δ α π τ π α δτ. Alternatively, we could substitute ν π α f to obtain δt π α f eiανt dν. The new limits in terms of ν are either ν if α > or else ν ± if α < and in the latter case we need to reverse the order of the limits and multiply by. Thus we end up with δt π α f eiανt dν which is the same result as before. 7. Xf e iπft dt δf. This follows from the answer to question 6 with α π. Solution Sheet 4 Page of

15 Ver 54 E. Fourier Series and Transforms 4 8. The Fourier transform of a periodic waveform is iust the complex Fourier series coefficients multiplied by delta functions at the appropriate positive and negative frequencies. So Xf 6δf + + 6δf + 4iδf + 4iδf. 9. The complex Fourier series coefficients are V n F.5t.5T δte iπf t dt F [ e iπf t] t F i.e. the same for all n. In fact, xt is equal to vt but iust written in a different way. So, from the theorem on page 6-8 of the notes, Xf n X nδf nf F n δf nf. Thus the Fourier transform of an impulse train with spacing F is another impulse train with spacing F.. a If vt Xt cos t, then V f 5 δf + π + xf V f, so xt 5 δt + π + 5 δt π. b xt cos f eiπft df e if + e if e iπft df 5 eiπt+ π f df + 5 eiπt π f df. Xf.5.5 e iπft dt iπf 5 δf π. So, from the duality theorem, 5 δt + π + 5 δt π. [ ] e iπft.5 t.5 sin πf iπf i sin πf. Xf e at e iπft dt e a iπft dt a iπf that the value of e a iπft is zero at t provided that a >. πf. [ e a iπft ] t a iπf a+iπf. Note. a xt cos t.5+.5 cost. The gains at these component frequencies are Y X i and Y X i +i.4.8i. It follows from phasors that yt +. cost +.4 sint. The Fourier transforms are Xf.5δf +.5δ f + π +.5δ f π and Y f δf +. +.i δ f + π +..i δ f π. Note that the positive frequency term, δ f π, is multiplied by Y X iπf while the negative frequency term, δ f + π, is multiplied by its complex conjugate, Y X iπf. b From question we know that Xf iπf+5. So it follows that Y f Xf Y X iπf iπf + 5 iπf + iπf + 5 iπf + We can put the given expression over a common denominator: Equating the numerator to gives c 4 and d 4. Hence yt c iπf+5 + d iπf+ iπfc+d+c+5d iπf+5iπf+. 4 e 5t e t t t <. 4. yt xτxt τdτ. The integrand is only non-zero when the arguments of both top-hat functions lie in the range ±.5. Thus we must have.5 < τ <.5 and also.5 < t τ <.5 t.5 < τ < t +.5. We can therefore write yt t+.5 min.5, t+.5 max.5, t.5 dτ.5 dτ t <.5 The integration range is dτ t. t.5 + t t < t t empty if t > and so we can write yt which also equals yt t t t > as requested. From the convolution theorem, Y f X f sin πf π f. 5. [B] An energy signal has finite energy: xt dt <. A power signal has infinite energy B but finite power: lim A,B B A A xt dt <. The answers are therefore a P, b P, c N, d N, e N, f N, g E, h E, i P, i E, i P. The final example has zero average power but is not an energy signal because it has infinite energy. Solution Sheet 4 Page of

16 Ver 54 E. Fourier Series and Transforms 4 6. a We substitute ω πf to obtain: Xω + ω + i δ ω π + 4 δ ω π 4 + ω + i δ ω + 8π π π δω 8π + 4πi δω + 8π δω 8π. + ω The final line is obtained using the scaling formula for delta functions: c δcx δx. Thus we see that in the angular-frequency version of the Fourier transform, any continuous functions of f remain the same amplitude but delta functions are multiplied by π. The inverse transform is given by xt Xωe iωt π dω; this can be obtained by changing the variable in the normal inverse transform from f to ω. b Xω is exactly the same as Xω but divided by π. So Xω π + ω + 8πi δω + 8π δω 8π. The inverse transform is the same as in the previous part but multiplied by π, i.e. xt π ˆ Xωe iωt dω. Solution Sheet 4 Page of