{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

solution0

# solution0 - ECE 101 Linear Systems Fall 2009 Problem Set#0...

This preview shows pages 1–4. Sign up to view the full content.

ECE 101 – Linear Systems, Fall 2009 Problem Set #0 Solutions September 24, 2009 (send comments/questions to [email protected]) 1.48 Using Euler’s relation, we have: z 0 = r 0 e 0 = r 0 cos θ 0 + jr 0 sin θ 0 = x 0 + jy 0 Then z 1 through z 5 are: (a) z 1 = x 0 - jy 0 (b) z 2 = q x 2 0 + y 2 0 (c) z 3 = - x 0 - jy 0 = - z 0 (d) z 4 = - x 0 + jy 0 (e) z 5 = x 0 + jy 0 = z 0 (recall, e is periodic with period 2 π ) Plots: 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
1.49 Once in polar form re , the magnitude is given by r and angle given by θ . (a) 2 e jπ/ 3 (b) 5 e (c) 5 2 e j 5 π/ 4 (d) 5 e jtan - 1 (4 / 3) = 5 e j 53 . 13 o (e) 8 e - (f) 4 2 e j 5 π/ 4 (g) 2 2 e - j 5 π/ 12 (h) e - j 2 π/ 3 (i) e jπ/ 6 Plot: 1.51 By Euler’s relation, we have (i) e = cos θ + j sin θ and (ii) e - = cos θ - j sin θ (a) Summing (i) + (ii) yields cos θ = 1 2 ( e + e - ). (b) Subtracting (i) - (ii) yields sin θ = 1 2 j ( e - e - ). 1.52 (c) z + z * = x + jy + x - jy = 2 x = 2 Re { z } (d) z - z * = x + jy - x + jy = 2 jy = 2 jIm { z } 2
1.53 (a) ( e z ) * = ( e x + jy ) * = ( e x e jy ) * = e x e - jy = e x - jy = e z * (c) | z | = re = | r | = re - = | z * | In words, taking the conjugate means flipping across the real axis; this negates the angle, but does not affect the magnitude.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}