I have no idea where to begin.... need to do 8-13 through 8-15 Use z-transform p
ID: 1799833 • Letter: I
Question
I have no idea where to begin.... need to do 8-13 through 8-15
Use z-transform pair 3 in Table 8-1 to establish z-transform pairs 4 and 5. (Hint: First write and differentiate both sides with respect to alpha) Use z-transform pair 3 in Table 8-1 to establish z-transform pairs 6 and 7. (Hint: Again first write Let alpha=jb and equate real and imaginary parts.) Use z-transform pair 3 in Table 8-1 to establish z-transform pairs 8 and 9. (Hint: The procedure is the same as in Problem 8-14. What should you let alpha be now?) Show that the z-transform of a*x(nT) is X(z/a). Use this result to show that entry 3 in Table B-1 follows from entry 2.Explanation / Answer
8-13) i solved this one in another post for you 8-14) (summation) e^(-anT)z^-n = 1/{1 - e^-aT*z^-1} now,putting a = jb (summation) e^(-jbnT)z^-n = 1/{1 - e^-jbT*z^-1} => Z{e^(-jbnT)} = 1/{1 - e^-jbT*z^-1} Now, both sides are complex numbers we calculate their real and imaginary parts and equate them respectively that is Z{ real{e^(-jbnT)} } = real{ 1/{1 - e^-jbT*z^-1} } --------- (1) and Z{ imaginary{e^(-jbnT)} } = imaginary{ 1/{1 - e^-jbT*z^-1} } -------- (2) now, e^-jbnT = cos(bnT) - j*sin(bnT) so, real{e^(-jbnT)} = cos(bnT) and, imag{e^(-bnT)} = -sin(bnT) also, 1/{1 - e^-jbT*z^-1} = {1 - e^jbT*z^-1}/{1 - e^-jbT*z^-1}*{1 - e^jbT*z^-1} = {1 - e^jbT*z^-1}/{1 - z^-2 - (e^jbT*z^-1 + e^-jbT*z^-1) } = {1 - (cos(bT) + j*sin(bT))z^-1}/{1 - z^-2 - (cos(bT) + j*sin(bT) + cos(bT) - j*sin(bT))z^-1) } = {1 - (cos(bT) + j*sin(bT))z^-1}/{1 - z^-2 - (2cos(bT))z^-1) } so, real{ 1/{1 - e^-jbT*z^-1} } = {1 - cos(bT)z^-1}/{1 - z^-2 - (2cos(bT)z^-1) } and, imag{ 1/{1 - e^-jbT*z^-1} } = {-sin(bT)z^-1}/{1 - z^-2 - (2cos(bT)z^-1) } putting these values into (1) and (2), (1) => Z{ real{e^(-jbnT)} } = real{ 1/{1 - e^-jbT*z^-1} } => Z{ cos(bnT) } = {1 - cos(bT)z^-1}/{1 - z^-2 - (2cos(bT)z^-1) } (2) => Z{ imaginary{e^(-jbnT)} } = imaginary{ 1/{1 - e^-jbT*z^-1} } => Z{ -sin(bnT) } = {-sin(bT)z^-1}/{1 - z^-2 - (2cos(bT)z^-1) } => Z{ sin(bnT) } = {sin(bT)z^-1}/{1 - z^-2 - (2cos(bT)z^-1) } 8-15) just put a = a + jb and do as in 8-14) 8-16) Z{a^n*x(nT)} = (summation) a^n*x(nT)*z^-n = (summation) x(nT)*(1/a)^-n*z^-n = (summation) x(nT)*(z/a)^-n define Z = z/a Z{a^n*x(nT)} = (summation) x(nT)*(z/a)^-n = (summation) x(nT)*(Z)^-n = X(Z) { by definition of z transform X(z) of x(t) } = X(z/a) => Z{a^n*x(nT)} = X(z/a) --------- (3) First part proved Now, from property 2 in table Z{1} = 1/(1-z^-1) => if x(nT) = 1 => X(z) = Z{x(t)} = 1/(1-z^-1) ------------- (4) now, put a=e^(-aT) in (3) Z{a^n*x(nT)} = X(z/a) => Z{(e^-aT)^n*x(nT)} = X(z/(e^-aT)) => Z{e^-anT*x(nT)} = X(z/(e^-aT)) Now, put x(nT) =1 and using result of (4) Z{e^-anT*x(nT)} = X(z/(e^-aT)) => Z{e^-anT} = 1/ ( 1 - (z/e^-aT)^-1) => Z{e^-anT} = 1/ ( 1 - ((e^aT)*z)^-1) => Z{e^-anT} = 1/ ( 1 - e^-aT*(z^-1)) hence, property 3 is proved