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Author: Peter
Date:2020-10-09
Location:FZU

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第一章

1-2 判斷下列序列是否為周期序列，若是，確定其周期

(1) x ( n ) = A c o s ( 3 π n 7 ? π 8 ) x(n)=Acos\big(\dfrac{3\pi n}{7}-\dfrac{\pi}{8}\big) x(n)=Acos(73πn??8π?)

∵ ω = 2 π N T = 3 π 7 \omega=\dfrac{2\pi N}{T}=\dfrac{3\pi}{7} ω=T2πN?=73π?

∴ T = 14 N 3 T=\dfrac{14N}{3} T=314N?

∵當N=3時，T=14為x(n)的最小周期

(2) x ( n ) = e j ( n 8 ? π ) x(n)=e^{j\big(\dfrac{n}{8}-\pi\big)} x(n)=ej(8n??π)

∵ x ( n ) = c o s ( n 8 ? π ) ? j s i n ( n 8 ? π ) x(n)=cos(\dfrac{n}{8}-\pi)-jsin(\dfrac{n}{8}-\pi) x(n)=cos(8n??π)?jsin(8n??π)

∵ ω = 2 π N T = 1 8 \omega=\dfrac{2\pi N}{T}=\dfrac{1}{8} ω=T2πN?=81?

∴ T = 16 π N T=16\pi N T=16πN

∵T中含有 π \pi π，T不能為整數

∴x(n)不是周期序列

1-3 系統框圖如下，已知邊界條件為y(-1)=0,分別求出以下輸入序列時的y(n),并畫出圖形,其中 y ( n ) = x ( n ) + 1 3 y ( n ? 1 ) y(n)=x(n)+\dfrac{1}{3}y(n-1) y(n)=x(n)+31?y(n?1)

(1) x ( n ) = δ ( n ) x(n)=\delta(n) x(n)=δ(n)

• 時域求解零輸入:
  ∵差分方程為 y ( n ) ? 1 3 y ( n ? 1 ) = x ( n ) y(n)-\dfrac{1}{3}y(n-1)=x(n) y(n)?31?y(n?1)=x(n)
  ∴特征方程為 α ? 1 3 = 0 \alpha-\dfrac{1}{3}=0 α?31?=0
  ∴ y z i ( n ) = C 1 α n = C 1 ( 1 3 ) n y_{zi}(n)=C_1\alpha^n=C_1(\dfrac{1}{3})^n yzi?(n)=C1?αn=C1?(31?)n
• 頻域求解零狀態:
  ∵由雙邊Z變換得 Y ( z ) ( 1 ? 1 3 z ? 1 ) = X ( z ) Y(z)(1-\dfrac{1}{3}z^{-1})=X(z) Y(z)(1?31?z?1)=X(z)
  ∴ Y ( z ) = X ( z ) 1 ? 1 3 z ? 1 = z z ? 1 3 X ( z ) Y(z)=\dfrac{X(z)}{1-\dfrac{1}{3}z^{-1}}=\dfrac{z}{z-\dfrac{1}{3}}X(z) Y(z)=1?31?z?1X(z)?=z?31?z?X(z)
  ∵ x ( n ) = δ ( n ) x(n)=\delta(n) x(n)=δ(n)
  ∴ Y ( z ) = z z ? 1 3 Y(z)=\dfrac{z}{z-\dfrac{1}{3}} Y(z)=z?31?z?
  ∴ y z s ( n ) = ( 1 3 ) n u ( n ) y_{zs}(n)=(\dfrac{1}{3})^nu(n) yzs?(n)=(31?)nu(n)
• 全回應=零輸入+零狀態
  ∴ y ( n ) = y z i ( n ) + y z s ( n ) = C 1 ( 1 3 ) n + ( 1 3 ) n u ( n ) y(n)=y_{zi}(n)+y_{zs}(n)=C_1(\dfrac{1}{3})^n+(\dfrac{1}{3})^nu(n) y(n)=yzi?(n)+yzs?(n)=C1?(31?)n+(31?)nu(n)
  ∵將 y ( ? 1 ) = 0 代 入 y ( n ) 得 C 1 = 0 y(-1)=0代入y(n)得C_1=0 y(?1)=0代入y(n)得C1?=0
  ∴ y z i ( n ) = 0 y_{zi}(n)=0 yzi?(n)=0
  ∴ y ( n ) = y z s ( n ) = ( 1 3 ) n u ( n ) y(n)=y_{zs}(n)=(\dfrac{1}{3})^nu(n) y(n)=yzs?(n)=(31?)nu(n)
• y ( n ) y(n) y(n)影像：
  

(2) x ( n ) = u ( n ) x(n)=u(n) x(n)=u(n)

• 時域求解零輸入:
  ∵由(1)知 y z i ( n ) = C 1 ( 1 3 ) n y_{zi}(n)=C_1(\dfrac{1}{3})^n yzi?(n)=C1?(31?)n
• 頻域求解零狀態:
  ∵ x ( n ) = u ( n ) x(n)=u(n) x(n)=u(n)
  ∴ Y ( z ) = z z ? 1 ? z z ? 1 3 = z z ( z ? 1 ) ( z ? 1 3 ) = z ( A z ? 1 + B z ? 1 3 ) Y(z)=\dfrac{z}{z-1}\cdot\dfrac{z}{z-\dfrac{1}{3}}=z\dfrac{z}{(z-1)(z-\dfrac{1}{3})}=z\bigg(\dfrac{A}{z-1}+\dfrac{B}{z-\dfrac{1}{3}}\bigg) Y(z)=z?1z??z?31?z?=z(z?1)(z?31?)z?=z(z?1A?+z?31?B?)
  ∵ A = z z ? 1 3 ∣ z = 1 = 3 2 B = z z ? 1 ∣ z = 1 3 = ? 1 2 A=\dfrac{z}{z-\dfrac{1}{3}}\bigg|_{z=1}=\dfrac{3}{2}\ \ \ \ \ \ \ B=\dfrac{z}{z-1}\bigg|_{z=\dfrac{1}{3}}=-\dfrac{1}{2} A=z?31?z?∣∣∣∣?z=1?=23? B=z?1z?∣∣∣∣?z=31??=?21?
  ∴ Y ( z ) = 3 2 ? z z ? 1 ? 1 2 ? z z ? 1 3 Y(z)=\dfrac{3}{2}\cdot\dfrac{z}{z-1}-\dfrac{1}{2}\cdot\dfrac{z}{z-\dfrac{1}{3}} Y(z)=23??z?1z??21??z?31?z?
  ∴ y ( n ) = y z i ( n ) + y z s ( n ) = C 1 ( 1 3 ) n + 3 2 u ( n ) ? 1 2 ( 1 3 ) n u ( n ) y(n)=y_{zi}(n)+y_{zs}(n)=C_1(\dfrac{1}{3})^n+\dfrac{3}{2}u(n)-\dfrac{1}{2}\big(\dfrac{1}{3}\big)^nu(n) y(n)=yzi?(n)+yzs?(n)=C1?(31?)n+23?u(n)?21?(31?)nu(n)
  ∵將 y ( ? 1 ) = 0 代 入 y ( n ) 得 C 1 = 0 y(-1)=0代入y(n)得C_1=0 y(?1)=0代入y(n)得C1?=0
  ∴ y z i ( n ) = 0 y_{zi}(n)=0 yzi?(n)=0
  ∴ y ( n ) = y z s ( n ) = [ 3 2 ? 1 2 ( 1 3 ) n ] u ( n ) y(n)=y_{zs}(n)=\bigg[\dfrac{3}{2}-\dfrac{1}{2}\big(\dfrac{1}{3}\big)^n\bigg]u(n) y(n)=yzs?(n)=[23??21?(31?)n]u(n)
• y ( n ) y(n) y(n)影像：
  

(3) x ( n ) = G N ( n ) N = 5 x(n)=G_N(n) \ \ \ \ N=5 x(n)=GN?(n) N=5

∵ x ( n ) = G 5 ( n ) = u ( n ) ? u ( n ? 5 ) x(n)=G_5(n)=u(n)-u(n-5) x(n)=G5?(n)=u(n)?u(n?5)
∵由(1)知 h ( n ) = ( 1 3 ) n u ( n ) h(n)=(\dfrac{1}{3})^nu(n) h(n)=(31?)nu(n)
∴ y ( n ) = h ( n ) ? x ( n ) = ∑ m = ? ∞ + ∞ h ( m ) x ( n ? m ) y(n)=h(n)*x(n)=\sum\limits_{m=-∞}^{+∞}h(m)x(n-m) y(n)=h(n)?x(n)=m=?∞∑+∞?h(m)x(n?m)
= ∑ m = ? ∞ + ∞ ( 1 3 ) m u ( m ) [ u ( n ? m ) ? u ( n ? m ? 5 ) ] \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{m=-∞}^{+∞}(\dfrac{1}{3})^mu(m)[u(n-m)-u(n-m-5)] =m=?∞∑+∞?(31?)mu(m)[u(n?m)?u(n?m?5)]
= ∑ m = 0 n ( 1 3 ) m ? ∑ m = 0 n ? 5 ( 1 3 ) m \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{m=0}^{n}(\dfrac{1}{3})^m-\sum\limits_{m=0}^{n-5}(\dfrac{1}{3})^m =m=0∑n?(31?)m?m=0∑n?5?(31?)m
= ∑ m = n ? 4 n ( 1 3 ) m 其 中 n ? 4 ≥ 0 \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{m=n-4}^{n}(\dfrac{1}{3})^m \ \ \ \ 其中n-4\geq0 =m=n?4∑n?(31?)m 其中n?4≥0
= ( 1 3 ) n ? 4 [ 1 ? ( 1 3 ) 5 ] 1 ? 1 3 u ( n ? 4 ) \ \ \ \ \ \ \ \ \ \ \ =\dfrac{(\dfrac{1}{3})^{n-4}[1-(\dfrac{1}{3})^5]}{1-\dfrac{1}{3}}u(n-4) =1?31?(31?)n?4[1?(31?)5]?u(n?4)
= 121 3 n n = 4 , 5... \ \ \ \ \ \ \ \ \ \ \ =\dfrac{121}{3^n} \ \ \ \ \ n=4,5... =3n121? n=4,5...
∵ y ( n ) = h ( n ) ? [ u ( n ) ? u ( n ? 5 ) ] , 當 n = 0 , 1 , 2 , 3 , 4 時 y ( n ) = h ( n ) ? u ( n ) 既 ( 2 ) 中 的 解 y(n)=h(n)*[u(n)-u(n-5)],當n=0,1,2,3,4時y(n)=h(n)*u(n)既(2)中的解 y(n)=h(n)?[u(n)?u(n?5)],當n=0,1,2,3,4時y(n)=h(n)?u(n)既(2)中的解
∴ y ( n ) = [ 3 2 ? 1 2 ( 1 3 ) n ] [ u ( n ) ? u ( n ? 5 ) ] + 121 3 n u ( n ? 5 ) y(n)=\bigg[\dfrac{3}{2}-\dfrac{1}{2}\big(\dfrac{1}{3}\big)^n\bigg][u(n)-u(n-5)]+\dfrac{121}{3^n}u(n-5) y(n)=[23??21?(31?)n][u(n)?u(n?5)]+3n121?u(n?5)

• y ( n ) y(n) y(n)影像：
  

1-4 已知線性時不變系統的單位沖激回應h(n)，輸入x(n)，求輸出序列y(n)，并畫出y(n)的圖形

(1) h ( n ) = G N ( n ) x ( n ) = G N ( n ) N = 4 h(n)=G_N(n) \ \ \ \ \ x(n)=G_N(n)\ \ \ \ N=4 h(n)=GN?(n) x(n)=GN?(n) N=4

∵ h ( n ) = x ( n ) = u ( n ) ? u ( n ? 4 ) h(n)=x(n)=u(n)-u(n-4) h(n)=x(n)=u(n)?u(n?4)

∴ h ( n ) = x ( n ) = { 1 , 1 , 1 , 1 } n = 0 , 1 , 2 , 3 h(n)=x(n)=\{1,1,1,1\} \ \ \ \ \ n=0,1,2,3 h(n)=x(n)={1,1,1,1} n=0,1,2,3

∴根據不進位乘法可得 y ( n ) = { 1 , 2 , 3 , 4 , 3 , 2 , 1 } n = 0 , 1...6 y(n)=\{1,2,3,4,3,2,1\} \ \ \ \ n=0,1...6 y(n)={1,2,3,4,3,2,1} n=0,1...6

y ( n ) y(n) y(n)影像:

(2) h ( n ) = 2 n G 4 ( n ) x ( n ) = δ ( n ) ? δ ( n ? 2 ) h(n)=2^nG_4(n) \ \ \ \ \ x(n)=\delta(n)-\delta(n-2) h(n)=2nG4?(n) x(n)=δ(n)?δ(n?2)

∵ h ( n ) = 2 n [ u ( n ) ? u ( n ? 4 ) ] h(n)=2^n[u(n)-u(n-4)] h(n)=2n[u(n)?u(n?4)]

∴ h ( n ) = { 1 , 2 , 4 , 8 } n = 0 , 1 , 2 , 3 h(n)=\{1,2,4,8\} \ \ \ \ \ \ n=0,1,2,3 h(n)={1,2,4,8} n=0,1,2,3

∵ x ( n ) = { 1 , 0 , ? 1 } n = 0 , 1 , 2 x(n)=\{1,0,-1\} \ \ \ \ \ \ \ n=0,1,2 x(n)={1,0,?1} n=0,1,2

∴根據不進位乘法可得 y ( n ) = { 1 , 2 , 3 , 6 , ? 4 , ? 8 } n = 0 , 1...5 y(n)=\{1,2,3,6,-4,-8\} \ \ \ \ n=0,1...5 y(n)={1,2,3,6,?4,?8} n=0,1...5

y ( n ) y(n) y(n)影像:

(3) h ( n ) = 0. 5 n u ( n ) x ( n ) = G 5 ( n ) h(n)=0.5^nu(n) \ \ \ \ \ x(n)=G_5(n) h(n)=0.5nu(n) x(n)=G5?(n)

∵ y ( n ) = x ( n ) ? h ( n ) = 0. 5 n u ( n ) ? [ u ( n ) ? u ( n ? 5 ) ] y(n)=x(n)*h(n)=0.5^nu(n)*[u(n)-u(n-5)] y(n)=x(n)?h(n)=0.5nu(n)?[u(n)?u(n?5)]
= ∑ m = ? ∞ + ∞ x ( m ) h ( n ? m ) \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{m=-∞}^{+∞}x(m)h(n-m) =m=?∞∑+∞?x(m)h(n?m)
= ∑ m = ? ∞ + ∞ 0. 5 m u ( m ) [ u ( n ? m ) ? u ( n ? m ? 5 ) ] \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{m=-∞}^{+∞}0.5^mu(m)[u(n-m)-u(n-m-5)] =m=?∞∑+∞?0.5mu(m)[u(n?m)?u(n?m?5)]
= ∑ m = 0 n ( 1 2 ) m ? ∑ m = 0 n ? 5 ( 1 2 ) m \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{m=0}^{n}(\dfrac{1}{2})^m-\sum\limits_{m=0}^{n-5}(\dfrac{1}{2})^m =m=0∑n?(21?)m?m=0∑n?5?(21?)m
= ∑ m = n ? 4 n ( 1 2 ) m 其 中 n ? 4 ≥ 0 \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{m=n-4}^{n}(\dfrac{1}{2})^m \ \ \ \ \ 其中n-4\geq0 =m=n?4∑n?(21?)m 其中n?4≥0
= ( 1 2 ) n ? 4 [ 1 ? ( 1 2 ) 5 ] 1 ? 1 2 \ \ \ \ \ \ \ \ \ \ \ =\dfrac{(\dfrac{1}{2})^{n-4}[1-(\dfrac{1}{2})^5]}{1-\dfrac{1}{2}} =1?21?(21?)n?4[1?(21?)5]?
= 31 2 n u ( n ? 4 ) \ \ \ \ \ \ \ \ \ \ \ =\dfrac{31}{2^n}u(n-4) =2n31?u(n?4)

∵ 當 n = 0 , 1 , 2 , 3 , 4 時 y ( n ) = 0. 5 n u ( n ) ? u ( n ) 當n=0,1,2,3,4時y(n)=0.5^nu(n)*u(n) 當n=0,1,2,3,4時y(n)=0.5nu(n)?u(n)
∵ Y ( z ) = 2 z z ? 1 ? z z ? 1 2 Y(z)=\dfrac{2z}{z-1}-\dfrac{z}{z-\dfrac{1}{2}} Y(z)=z?12z??z?21?z?
∴ y ( n ) = 2 ? ( 1 2 ) n n = 0 , 1 , 2 , 3 , 4 y(n)=2-(\dfrac{1}{2})^n \ \ \ \ \ n=0,1,2,3,4 y(n)=2?(21?)n n=0,1,2,3,4

∴ y ( n ) = [ 2 ? ( 1 2 ) n ] [ u ( n ) ? u ( n ? 5 ) ] + 31 2 n u ( n ? 5 ) y(n)=[2-(\dfrac{1}{2})^n][u(n)-u(n-5)]+\dfrac{31}{2^n}u(n-5) y(n)=[2?(21?)n][u(n)?u(n?5)]+2n31?u(n?5)

y ( n ) y(n) y(n)影像:

1-7 下示每個系統中，x(n)表示激勵、y(n)表示回應，判斷系統是否線性，是否時變？

(1) y ( n ) = x ( n ) s i n ( 2 π 7 n + π 6 ) y(n)=x(n)sin(\dfrac{2\pi}{7}n+\dfrac{\pi}{6}) y(n)=x(n)sin(72π?n+6π?)

∵ T [ C 1 x 1 ( n ) + C 2 x 2 ( n ) ] = [ C 1 x 1 ( n ) + C 2 x 2 ( n ) ] s i n ( 2 π 7 n + π 6 ) T[C_1x_1(n)+C_2x_2(n)]=[C_1x_1(n)+C_2x_2(n)]sin(\dfrac{2\pi}{7}n+\dfrac{\pi}{6}) T[C1?x1?(n)+C2?x2?(n)]=[C1?x1?(n)+C2?x2?(n)]sin(72π?n+6π?)

∵ [ C 1 y 1 ( n ) + C 2 y 2 ( n ) ] = [ C 1 x 1 ( n ) + C 2 x 2 ( n ) ] s i n ( 2 π 7 n + π 6 ) [C_1y_1(n)+C_2y_2(n)]=[C_1x_1(n)+C_2x_2(n)]sin(\dfrac{2\pi}{7}n+\dfrac{\pi}{6}) [C1?y1?(n)+C2?y2?(n)]=[C1?x1?(n)+C2?x2?(n)]sin(72π?n+6π?)

∴ T [ C 1 x 1 ( n ) + C 2 x 2 ( n ) ] = [ C 1 y 1 ( n ) + C 2 y 2 ( n ) ] T[C_1x_1(n)+C_2x_2(n)]={[C_1y_1(n)+C_2y_2(n)]} T[C1?x1?(n)+C2?x2?(n)]=[C1?y1?(n)+C2?y2?(n)]

∵ T [ x ( n ? m ) ] = x ( n ? m ) s i n ( 2 π 7 n + π 6 ) T[x(n-m)]=x(n-m)sin(\dfrac{2\pi}{7}n+\dfrac{\pi}{6}) T[x(n?m)]=x(n?m)sin(72π?n+6π?)

∵ y ( n ? m ) = x ( n ? m ) s i n [ 2 π 7 ( n ? m ) + π 6 ] y(n-m)=x(n-m)sin[\dfrac{2\pi}{7}(n-m)+\dfrac{\pi}{6}] y(n?m)=x(n?m)sin[72π?(n?m)+6π?]

∵ T [ x ( n ? m ) ] =? y ( n ? m ) T[x(n-m)]\not ={y(n-m)} T[x(n?m)]?=y(n?m)

∴系統為線性時變

(2) y ( n ) = [ x ( n ) ] 2 y(n)=[x(n)]^2 y(n)=[x(n)]2

∵ T [ C 1 x 1 ( n ) + C 2 x 2 ( n ) ] = [ C 1 x 1 ( n ) + C 2 x 2 ( n ) ] 2 T[C_1x_1(n)+C_2x_2(n)]=[C_1x_1(n)+C_2x_2(n)]^2 T[C1?x1?(n)+C2?x2?(n)]=[C1?x1?(n)+C2?x2?(n)]2

∵ [ C 1 y 1 ( n ) + C 2 y 2 ( n ) ] = [ C 1 x 1 ( n ) ] 2 + [ C 2 x 2 ( n ) ] 2 [C_1y_1(n)+C_2y_2(n)]=[C_1x_1(n)]^2+[C_2x_2(n)]^2 [C1?y1?(n)+C2?y2?(n)]=[C1?x1?(n)]2+[C2?x2?(n)]2

∴ T [ C 1 x 1 ( n ) + C 2 x 2 ( n ) ] =? [ C 1 y 1 ( n ) + C 2 y 2 ( n ) ] T[C_1x_1(n)+C_2x_2(n)]\not={[C_1y_1(n)+C_2y_2(n)]} T[C1?x1?(n)+C2?x2?(n)]?=[C1?y1?(n)+C2?y2?(n)]

∵ T [ x ( n ? m ) ] = [ x ( n ? m ) ] 2 T[x(n-m)]=[x(n-m)]^2 T[x(n?m)]=[x(n?m)]2

∵ y ( n ? m ) = [ x ( n ? m ) ] 2 y(n-m)=[x(n-m)]^2 y(n?m)=[x(n?m)]2

∵ T [ x ( n ? m ) ] = y ( n ? m ) T[x(n-m)]={y(n-m)} T[x(n?m)]=y(n?m)

∴系統為非線性時不變

1-9 用卷積法求系統回應y(n)

(1) y ( n ) = a n u ( n ) 0 < a < 1 , h ( n ) = β n u ( n ) 0 < β < 1 , β =? a y(n)=a^nu(n) \ \ \ 0<a<1,h(n)=\beta^nu(n) \ \ \ 0<\beta<1,\beta\not ={a} y(n)=anu(n) 0<a<1,h(n)=βnu(n) 0<β<1,β?=a

∵ y ( n ) = x ( n ) ? h ( n ) = ∑ m = ? ∞ + ∞ x ( m ) h ( n ? m ) y(n)=x(n)*h(n)=\sum\limits_{m=-∞}^{+∞}x(m)h(n-m) y(n)=x(n)?h(n)=m=?∞∑+∞?x(m)h(n?m)
∴ y ( n ) = ∑ m = ? ∞ + ∞ α m u ( m ) β ( n ? m ) u ( n ? m ) y(n)=\sum\limits_{m=-∞}^{+∞}\alpha^mu(m)\beta^{(n-m)}u(n-m) y(n)=m=?∞∑+∞?αmu(m)β(n?m)u(n?m)
= ∑ m = 0 n α m β ( n ? m ) \ \ \ \ \ \ \ \ \ \ \ =\sum\limits_{m=0}^{n}\alpha^m\beta^{(n-m)} =m=0∑n?αmβ(n?m)
= β n ∑ m = 0 n ( α β ) m \ \ \ \ \ \ \ \ \ \ \ =\beta^n\sum\limits_{m=0}^{n}\big(\dfrac{\alpha}{\beta}\big)^m =βnm=0∑n?(βα?)m
= β n ? 1 ? ( α β ) n + 1 1 ? α β \ \ \ \ \ \ \ \ \ \ \ =\beta^n\cdot\dfrac{1-\big(\dfrac{\alpha}{\beta}\big)^{n+1}}{1-\dfrac{\alpha}{\beta}} =βn?1?βα?1?(βα?)n+1?
= β n + 1 ? α n + 1 β ? α u ( n ) \ \ \ \ \ \ \ \ \ \ \ =\dfrac{\beta^{n+1}-\alpha^{n+1}}{\beta-\alpha}u(n) =β?αβn+1?αn+1?u(n)

(2) x ( n ) = u ( n ) , h ( n ) = δ ( n ? 2 ) ? δ ( n ? 3 ) x(n)=u(n),h(n)=\delta(n-2)-\delta(n-3) x(n)=u(n),h(n)=δ(n?2)?δ(n?3)

∵ y ( n ) = x ( n ) ? h ( n ) y(n)=x(n)*h(n) y(n)=x(n)?h(n)
∵ x ( n ) ? δ ( n ? m ) = x ( n ? m ) x(n)*\delta(n-m)=x(n-m) x(n)?δ(n?m)=x(n?m)
∴ y ( n ) = u ( n ? 2 ) ? u ( n ? 3 ) = δ ( n ? 2 ) y(n)=u(n-2)-u(n-3)=\delta(n-2) y(n)=u(n?2)?u(n?3)=δ(n?2)