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Solutions to Exercises

3.10 Solutions to Exercises

1. We apply the sampling property of the G t function for all expressions except (e) where we apply
the sifting property. For part (f) we apply the sampling property of the doublet.

We recall that the sampling property states that f t G t – a = fa G t – a . Thus,

§
§
§
---G t –
a. sin tG t – S· --- = sin t G t – S· --- = sin S § S· --- = 0.5G t – S· ---
© 6 ¹ t = S 6 © e 6 ¹ 6 © 6 ¹ © 6 ¹
§
§
b. cos 2tG t – S· --- = cos 2t G t – S· --- = cos S § S · --- = 0
---G t –
© 4 ¹ t = S 4 © e 4 ¹ 2 © 4 ¹
§
§
c. cos 2 tG t – S· --- = 1 - cos 2t G t – S· --- = 1 - cos S G § t – S· --- = 1 - G § t – S· --- = 0
-- 1 +
-- 11–
-- 1 +
© 2 ¹ 2 © 2 ¹ 2 © 2 ¹ 2 © 2 ¹
t = S 2
e
§
§
§
---G t –
d. tan 2tG t – S· --- = tan 2t G t – S· --- = tan S § S · --- = G t – S· ---
© 8 ¹ t = S 8 © e 8 ¹ 4 © 8 ¹ © 8 ¹
f
We recall that the sampling property states that ³ ft G t – D t = f D . Thus,
d

– f
f 2 – t 2 – t
e. ³ t e G t – 2 t = t e t = 2 = 4e – 2 = 0.54
d

– f
We recall that the sampling property for the doublet states that
f t G' t – a = fa G' t – a – f ' a G t – a

Thus,
2 1 § S · 2 1 § S · d 2 § S ·
sin tG t – --- = sin t G t – --- – ----- sin t G t – ---
© 2 ¹ t = S 2 © e 2 ¹ dt t = S 2 © e 2 ¹
1 1 § S · § S ·
f. = -- 1 – - cos 2t G t – --- – sin 2t G t – ---
2 t = S 2 © e 2 ¹ t = S 2 © e 2 ¹
1 1 § S · S · 1 § S ·
§
-- 1 + = - 1 G t – --- – sin SG t – --- = G t – ---
2 © 2 ¹ © 2 ¹ © 2 ¹
2.
vt = e – 2t > u t – u t – 2 @ + 10t – 30 u t – 2 – u t – 3 @

>
a. 0 0 0 0
+ – 10t + 50 u t – > 0 3 – u t – 5 @ + 10t – 70 u t – 5 – u t – 7 @

>

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Circuit Analysis II with MATLAB Applications 3-21
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