M. J. Roberts - 7/12/03 Chapt er 2 - Mathematical Description of Signals Solutions 1. If g(t)= 7e−2t−3 write out and simplify (a) g(3)= 7e−9 (b) g(2−t)= 7e−2(2−t)−3 = 7e−7+2t (c) g t +4 = 7e−5t−11   10 ( ) (d) g jt = 7e−j2t−3 ( ) ( ) (e) g jt +g −jt = 7e−3 e−j2t +ej2t = 7e−3cos(2t) 2 2  jt−3 −jt−3 g  +g  (f)  2   2  = 7e−jt +ejt = 7cos(t) 2 2 2. If g(x)= x2−4x +4 write out and simplify (a) g(z)= z2−4z+4 (b) g(u+v)=(u+v)2−4(u+v)+4 = u2 +v2 +2uv−4u−4v +4 ( ) ( ) ( ) (c) g ejt = ejt 2−4ejt +4 =ej2t −4ejt +4 = ejt −2 2 (d) g(g(t))=g(t2 −4t+4)=(t2 −4t+4)2−4(t2−4t+4)+4 g(g(t))= t4 −8t3 +20t2 −16t+4 (e) g(2)= 4−8+4 =0 3. What would be the numerical value of “g” in each of the following MATLAB instructions? (a) t = 3 ; g = sin(t) ; 0.1411 (b) x = 1:5 ; g = cos(pi*x) ; [-1,1,-1,1,-1] (c) f = -1:0.5:1 ; w = 2*pi*f ; g = 1./(1+j*w) ; Solutions 2-1 M. J. Roberts - 7/12/03 0.0247+ j0.155   0.0920+ j0.289     1   0.0920− j0.289 0.0247− j0.155 4. Let two functions be defined by 1 , sin(20πt)≥0 t , sin(2πt)≥0 x (t)= and x (t)= . 1 −1 , sin(20πt)<0 2 −t , sin(2πt)<0 Graph the product of these two functions versus time over the time range, −2< t<2. x(t) 2 t -2 2 -2 5. For each function, g(t), sketch g(−t), −g(t), g(t−1), and g(2t). (a) (b) g(t) g(t) 4 3 t -1 t 2 1 -3 g(-t) g(-t) -g(t) -g(t) 4 3 3 t -1 t t 1 t -2 1 2 -1 -3 4 -3 g(t-1) g(t-1) g(2t) g(2t) 4 3 4 3 -1 2 t t t t 1 3 1 2 1 1 2 -3 -3 ( ) 6. A function, G f , is defined by Solutions 2-2 M. J. Roberts - 7/12/03 ( )  f  G f =e−j2πf rect  .   2 ( ) ( ) Graph the magnitude and phase of G f −10 +G f +10 over the range, −20< f <20. G(f −10)+G(f +10)=e−j2π(f−10)rect f −10 +e−j2π(f+10)rect f +10     2 2 |G( f )| 1 f -20 20 Phase of G( f ) π f -20 20 -π 7. Sketch the derivatives of these functions. (All sketches at end.) (a) g(t)=sinc(t) g′(t)= π2tcos(πt)−πsin(πt) = πtcos(πt)−sin(πt) (πt)2 πt2 (b) g(t)=(1−e−t)u(t) g′(t)=e−t , t≥0=e−tu(t) 0 , t<0 (a) (b) x(t) x(t) 1 1 t t -4 4 -1 4 -1 -1 dx/dt dx/dt 1 1 t t -4 4 -1 4 -1 -1 8. Sketch the integral from negative infinity to time, t, of these functions which are zero for all time before time, t=0. Solutions 2-3 M. J. Roberts - 7/12/03 g(t) g(t) 1 1 1 2 3 t t 1 1 2 3 2 ∫ g(t) dt ∫ g(t) dt 1 1 1 2 t t 1 2 3 1 2 3 9. Find the even and odd parts of these functions. (a) g(t)=2t2−3t+6 g (t)= 2t2 −3t+6+2(−t)2 −3(−t)+6 = 4t2 +12 =2t2 +6 e 2 2 g (t)= 2t2 −3t+6−2(−t)2 +3(−t)−6 = −6t =−3t o 2 2  π (b) g(t)=20cos40πt−    4  π  π 20cos40πt−  +20cos−40πt−      g (t)= 4 4 e 2 ( ) ( ) ( ) ( ) ( ) Using cos z +z =cos z cos z −sin z sin z 1 2 1 2 1 2    π  π  20cos(40πt)cos−  −sin(40πt)sin−         4 4     +20cos(−40πt)cos−π −sin(−40πt)sin−π g (t)=   4 4  e 2   π π  20cos(40πt)cos  +sin(40πt)sin         4 4     +20cos(40πt)cosπ −sin(40πt)sinπ g (t)=   4 4  e 2 Solutions 2-4 M. J. Roberts - 7/12/03 g (t)=20cosπcos(40πt)= 20 cos(40πt) e 4 2  π  π 20cos40πt−  −20cos−40πt−      g (t)= 4 4 o 2 ( ) ( ) ( ) ( ) ( ) Using cos z +z =cos z cos z −sin z sin z 1 2 1 2 1 2    π  π  20cos(40πt)cos−  −sin(40πt)sin−         4 4     −20cos(−40πt)cos−π −sin(−40πt)sin−π g (t)=   4 4  o 2   π π  20cos(40πt)cos  +sin(40πt)sin         4 4     −20cos(40πt)cosπ −sin(40πt)sinπ g (t)=   4 4  o 2 g (t)=20sinπsin(40πt)= 20 sin(40πt) o 4 2 (c) g(t)= 2t2 −3t+6 1+t 2t2 −3t+6 2t2 +3t+6 + g (t)= 1+t 1−t e 2 (2t2−3t+6)(1−t)+(2t2 +3t+6)(1+t) g (t)= (1+t)(1−t) e 2 g (t)= 4t2 +(12+)6t2 = 6+5t2 e 21−t2 1−t2 2t2 −3t+6 2t2 +3t+6 − g (t)= 1+t 1−t o 2 Solutions 2-5 M. J. Roberts - 7/12/03 (2t2−3t+6)(1−t)−(2t2 +3t+6)(1+t) g (t)= (1+t)(1−t) o 2 g (t)= −6t−(4t3 −)12t =−t2t2 +9 o 21−t2 1−t2 sin(πt) sin(−πt) (d) g(t)=sinc(t) g (t)= πt + −πt = sin(πt) g (t)=0 e 2 πt o (e) g(t)= t(2−t2)(1+4t2) g(t)= {t (2−t2)(1+4t2) 12314243 odd even even Therefore g(t) is odd, g (t)=0 and g (t)= t(2−t2)(1+4t2) e o (f) g(t)= t(2−t)(1+4t) g (t)= t(2−t)(1+4t)+(−t)(2+t)(1−4t) g (t)= 7t2 e 2 e g (t)= t(2−t)(1+4t)−(−t)(2+t)(1−4t) g (t)= t(2−4t2) o 2 o 10.Sketch the even and odd parts of these functions. Solutions 2-6 M. J. Roberts - 7/12/03 g(t) g(t) 1 1 t t 1 1 2 -1 g (t) g (t) e e 1 1 t t 1 1 2 -1 g (t) g (t) o o 1 1 t t 1 1 2 -1 (a) (b) ( ) 11.Sketch the indicated product or quotient, g t , of these functions. (a) (b) 1 1 -1 -1 t t 1 1 -1 -1 g(t) g(t) 1 1 Multiplication Multiplication -1 t t -1 1 1 -1 g(t) g(t) 1 1 -1 t t 1 -1 1 -1 -1 Solutions 2-7 M. J. Roberts - 7/12/03 (c) (d) 1 1 t t -1 1 g(t) g(t) Multiplication Multiplication 1 1 t t 1 1 g(t) g(t) -1 1 t -1 1 t -1 1 (e) (f) 1 ... 1 ... t 1 t -1 1 -1 g(t) -1 g(t) 1 1 Multiplication t Multiplication 1 t -1 -1 1 g(t) g(t) 1 ... ... 1 t -1 1 t 1 -1 -1 (g) (h) 1 1 t t -1 -1 -1 1 g(t) g(t) π Division Division 1 t t 1 1 g(t) g(t) 1 t -1 -1 1 t 12. Use the properties of integrals of even and odd functions to evaluate these integrals in the quickest way. Solutions 2-8 M. J. Roberts - 7/12/03 1 1 1 1 (a) ∫(2+t)dt= ∫ 2{ dt+ ∫ {t dt=2∫2dt= 4 −1 −1even −1odd 0 (b) 1 1 1 1 2∫0 [4cos(10πt)+8sin(5πt)]dt= 2∫0 4cos(10πt)dt+ 2∫0 8sin(5πt)dt=82∫0cos(10πt)dt= 8 14243 14243 10π −1 −1 even −1 odd 0 20 20 20 1 20 (c) ∫ 4 t{ cos(10πt)dt=0 14243 1 o1dd4243 − even 20 odd 1 1  1  1 (d) 1∫0 t{ s1i4n(2104π3t)dt=21∫0tsin(10πt)dt=2−tcos1(01π0πt)10 +1∫0cos1(01π0πt)dt −1 1odd42od4d 3 0  0 0  10 even 1  1  1∫0 t{ s1i4n(2104π3t)dt==21010π+ si(n1(01π0)π2t)10= 501π −1 1odd42od4d 3  0  10 even 1 1 1 [ ] ( ) (e) ∫e{−t dt=2∫e−tdt=2∫e−tdt=2 −e−t 1 =21−e−1 ≈1.264 0 −1even 0 0 1 (f) ∫ t{ e{−t dt=0 123 −1oddeven odd 13.Find the fundamental period and fundamental frequency of each of these functions. (a) g(t)=10cos(50πt) f =25 Hz , T = 1 s 0 0 25 (b) g(t)=10cos50πt+π f =25 Hz , T = 1 s  4 0 0 25 (c) g(t)=cos(50πt)+sin(15πt)  15 1 f =GCD25,  =2.5Hz , T = =0.4 s 0  2 0 2.5 (d) g(t)=cos(2πt)+sin(3πt)+cos5πt− 3π   4 Solutions 2-9 M. J. Roberts - 7/12/03  3 5 1 1 f =GCD1, ,  = Hz , T = =2s 0  2 2 2 0 1 2 ( ) 14. Find the fundamental period and fundamental frequency of g t . g(t) ... ... t 1 (a) ... ... (b) + g(t) t 1 ... ... t 1 ... ... t 1 (c) + g(t) ... ... t 1 1 (a) f = 3Hz and T = s 0 0 3 (b) f =GCD(6,4) =2Hz and T = 1 s 0 0 2 (c) f =GCD(6,5) =1Hz and T =1s 0 0 15.Plot these DT functions. [ ] 2πn 2π(n−2) (a) x n = 4cos  −3sin  , −24≤ n <24  12   8  x[n] 7 n -24 24 -7 n [ ] − (b) x n = 3ne 5 , −20≤ n<20 x[n] 6 n -20 20 -6 [ ] n2 (c) x n =21  +14n3 , −5≤ n<5   2 Solutions 2-10

Signals and Systems: Analysis of Signals Through Linear Systems - Solution manual

Similar Signals and Systems: Analysis of Signals Through Linear Systems - Solution manual