Question

5. (4 pts) Let X(ej) be the DTFT of a signal x[n] which is known to be zero for n < 0 and n > 3. We know X(eja) for four values of N as follows. X(@j0) = 10, X(eja/2) = 5 – 5j, X(ejt) = 0, X(ej37/2) = 5 + 5j (a) (3 pts) Find x[n]. (Hint: Compute the IDFT) (b) (1 pts) Find X(ej?).

Answer 1

0 Data → givan Xleto) 10 xV) 5-oj x()=0 x ( c1 217) = 5+5j a ele know DTFT oH signeel ain x levien -Jun Kin] ، له x(1) ** + 2 jun ju2 Julxo -*1-2]e + xfije + x[oje + K13 e ejn + a[2] e Since it is mentioned incit tor and any gen] no then x cair) .Ju X Loge evonxo + x2] - ej uz + X[2] € 7 X[3] :3ju punt zao x420) x[o] + Kei) + X[2] + x[s] Xco] + X[1] + X[2] + x3] =10

2 put u= / X(e) 5-58 -J1/2 x[] tie x61] + x(] 27192 + [3] e 3/2 X[0] -jx[i] X [2] + j x[j] = 5-15 X[o] - X[2] =j[ x[i] x[3] = 5-js port by comparing real and smaginary X[o] -*[2] XL) - X[3] = 5 put na T -JT X[ 2*) = 0 - J2 x[o] + an] e + X[2] e +86] eJ3T 0 Ex[0] * [] +X[2] - X13] x[•] - x[i] + x[2] - X[3] 2 - 131 xr put a = 3 -J399/2 Xléstyn] = 55= x[O] + XG] ē + *[2]e to x[3) Wys

5+5) = x[o] +jx2] + X[e] -j x[3] 5+5j x[o] -x [e] + j [Y217 x [37] from Here X[o] -X(2) 5 ] XL] - x[3] 5 then add Eg? 3 & 1 gel 10 EXCO] + 2x[?] X[o] + x[?] 9 xľo] -y[e] eqin XLOJ +X[21 x [] -*[2] + 5 2 x[0] 10 X [0] = 10 -5 2 X[0] = 5 and R2) = 0 subs freete En € & Chin x [of + x[1] +X50] +X[3] =10. x80] XC,3 +k[:] =*[0] = 2X[i] + 2*[3] =10 0 +

کی = [s] + [۱] * -foom Eqn [1] -< [3] = 5 زارا by adding <rig and ¿vid we ged 2 [] 10 [1] 5 aay (3) ہ « (۱] = 5 ] = 0 finarny منا ged 5 = [ه ] 6 = [2] * ا ت [] * ه: [لا * [] S 5 T | | * [] = بها 0 5 3 } b) ( لغ ) * 27 لة [7th 3 'h: -

x [eling Juno xlo) e +x6,78u7x cering X (3) Bju 5*1 + 5e -Ja :yu toxē+ oxe ejer -juz x[dir] = = 5+ hone