1.Assume x ( t ) ↔ F T X ( j ω ) x(t) \xleftrightarrow[]{FT}X(j\omega) x(t)FT ​X(jω), discuss the following problems:

1. Can e j a ω X ( ω ) e^{ja \omega}X(\omega) ejaωX(ω) be real, how?
2. Can X ( j ω ) X(j\omega) X(jω) be periodic, how?
3. Can ∫ − ∞ ∞ ω X ( j ω ) d ω \int_{-\infty}^\infty{\omega X(j \omega) d\omega} ∫−∞∞​ωX(jω)dω be zero, how?
4. Can R e ( X ( j ω ) ) = 0 Re ({{{X(j\omega)}}})=0 Re(X(jω))=0 or I m ( X ( j ω ) ) = 0 Im(X{(j\omega)})=0 Im(X(jω))=0 , how?

Solve:
1). 根据题设有：
x ( t ) ↔ F T X ( j ω ) (1-1) x(t) \xleftrightarrow[]{FT}X(j\omega) \tag{1-1} x(t)FT ​X(jω)(1-1)
结合FT的时移性质，有：
x ( t + a ) ↔ F T e j a ω X ( ω ) = X 1 ( j ω ) (1-2) x(t+a) \xleftrightarrow[]{FT} e^{ja \omega}X(\omega) =X_{1}(j\omega)\tag{1-2} x(t+a)FT ​ejaωX(ω)=X1​(jω)(1-2)
根据尺度变换特性及共轭对称性,若：
x ( t ) ↔ F T X ( j ω ) (1-3) x(t) \xleftrightarrow[]{FT}X(j\omega) \tag{1-3} x(t)FT ​X(jω)(1-3)
则有：
x ( − t ) ↔ F T X ( − j ω ) (1-4) x(-t) \xleftrightarrow[]{FT}X(-j\omega) \tag{1-4} x(−t)FT ​X(−jω)(1-4)
x ∗ ( t ) ↔ F T X ∗ ( − j ω ) (1-5) x^{*}(t) \xleftrightarrow[]{FT}X^{*}(-j\omega) \tag{1-5} x∗(t)FT ​X∗(−jω)(1-5)
可以看出，若要求 X 1 ( j ω ) X_{1}(j\omega) X1​(jω) 为实，则需要有：
X ∗ ( j ω ) = X ( j ω ) (1-6) X^{*}(j\omega)=X(j\omega) \tag{1-6} X∗(jω)=X(jω)(1-6)
反映在时域上即为：
x ( t + a ) = x ∗ ( t + a ) (1-7) x(t+a)=x^{*}(t+a) \tag{1-7} x(t+a)=x∗(t+a)(1-7)
x ( t + a ) = x ( − t + a ) (1-8) x(t+a)=x(-t+a) \tag{1-8} x(t+a)=x(−t+a)(1-8)
即要求 x ( t ) x(t) x(t) 为实信号，并且关于 t = a t=a t=a 对称

2).根据对偶性，若：
x ( t ) ↔ F T X ( j ω ) (2-1) x(t) \xleftrightarrow[]{FT}X(j\omega) \tag{2-1} x(t)FT ​X(jω)(2-1)
则有：
X ( t ) ↔ F T 2 π x ( − j ω ) (2-2) X(t) \xleftrightarrow[]{FT}2\pi x(-j\omega) \tag{2-2} X(t)FT ​2πx(−jω)(2-2)
根据所学，周期信号的FT为频域的冲激串，即若 x ( t ) x(t) x(t) 为周期信号，则：

x ( t ) ↔ F T X ( j ω ) = 2 π ∑ k = − ∞ + ∞ a k δ ( ω − k ω 0 ) (2-3) x(t) \xleftrightarrow[]{FT}X(j\omega) \tag{2-3}=2 \pi \sum_{k=-\infty}^{+\infty} a_{k} \delta(\omega- k\omega_{0}) x(t)FT ​X(jω)=2πk=−∞∑+∞​ak​δ(ω−kω0​)(2-3)

则根据对偶性：
X ( t ) ↔ F T 2 π x ( − j ω ) (2-4) X(t) \xleftrightarrow[]{FT}2\pi x(-j\omega) \tag{2-4} X(t)FT ​2πx(−jω)(2-4)
x ( − j ω ) x(-j\omega) x(−jω) 即为频域上的周期信号。

综上：要求时域上的信号为冲激串形式，才可以保证其FT为周期信号。

3). 根据IFT：
x ( t ) = 1 2 π ∫ − ∞ ∞ e j ω t X ( j ω ) d ω (3-1) x(t)= \frac{1}{2\pi} \int_{-\infty}^\infty{e^{j\omega t}X(j \omega) d\omega} \tag{3-1} x(t)=2π1​∫−∞∞​ejωtX(jω)dω(3-1)

两边同时对 t t t 求导有：
x ′ ( t ) = ∫ − ∞ ∞ j ω X ( j ω ) e j ω t d ω (3-2) x^{'}(t)=\int_{-\infty}^\infty{j\omega X(j \omega)e^{j\omega t} d\omega} \tag{3-2} x′(t)=∫−∞∞​jωX(jω)ejωtdω(3-2)
令 t = 0 t=0 t=0 ,有：
x ′ ( 0 ) = ∫ − ∞ ∞ j ω X ( j ω ) d ω (3-3) x^{'}(0)=\int_{-\infty}^\infty{j\omega X(j \omega )d\omega} \tag{3-3} x′(0)=∫−∞∞​jωX(jω)dω(3-3)
令其为0，即为 x ′ ( t ) = 0 x^{'}(t)=0 x′(t)=0

4).根据共轭对称性，若要求 X ( j ω ) X(j\omega) X(jω) 实部为0，即为要求 X ( j ω ) X(j\omega) X(jω) 为纯虚数，推导如下：
若：
x ( t ) ↔ F T X ( j ω ) (3-4) x(t) \xleftrightarrow[]{FT}X(j\omega) \tag{3-4} x(t)FT ​X(jω)(3-4)
则有：
x ( − t ) ↔ F T X ( − j ω ) (3-5) x(-t) \xleftrightarrow[]{FT}X(-j\omega) \tag{3-5} x(−t)FT ​X(−jω)(3-5)
x ∗ ( t ) ↔ F T X ∗ ( − j ω ) (3-6) x^{*}(t) \xleftrightarrow[]{FT}X^{*}(-j\omega) \tag{3-6} x∗(t)FT ​X∗(−jω)(3-6)

现在要求： − X ( j ω ) = X ∗ ( j ω ) (3-7) -X(j\omega)=X^{*}(j\omega) \tag{3-7} −X(jω)=X∗(jω)(3-7)
反映在时域上为：
x ( t ) = x ∗ ( t ) (3-8) x(t)=x^{*}(t) \tag{3-8} x(t)=x∗(t)(3-8)
x ( t ) = − x ( − t ) (3-9) x(t)=-x(-t) \tag{3-9} x(t)=−x(−t)(3-9)
即要求 x ( t ) x(t) x(t) 为实奇信号。

同理，对于 X ( j w ) X(jw) X(jw) 虚部为0，有：

X ( j ω ) = X ∗ ( j ω ) (3-10) X(j\omega)=X^{*}(j\omega) \tag{3-10} X(jω)=X∗(jω)(3-10)
x ( t ) = x ∗ ( t ) (3-11) x(t)=x^{*}(t) \tag{3-11} x(t)=x∗(t)(3-11)
x ( t ) = x ( − t ) (3-12) x(t)=x(-t) \tag{3-12} x(t)=x(−t)(3-12)
即要求 x ( t ) x(t) x(t) 为实偶信号。

2
a) A sound signal x ( t ) x(t) x(t) is broadcasted in a reverberating room, which can be modeled as a system with the impulse response of h ( t ) = ∑ k = 0 + ∞ e − k T δ ( t − k T ) h(t)= \sum_{k=0}^{+\infty} e ^{-kT} \delta(t- kT) h(t)=∑k=0+∞​e−kTδ(t−kT) . Design a system that can recover x ( t ) x(t) x(t) from the reverberated sound signal.

b) When the room impulse response is modeled as h ( t ) = ∑ k = 0 K a k δ ( t − k T ) h(t)= \sum_{k=0}^{K} a_{k} \delta(t- kT) h(t)=∑k=0K​ak​δ(t−kT) , can we design a causal system that can recover x ( t ) x(t) x(t) from the reverberated sound signal? Specify the reason.

c) Summarize time-domain and frequency-domain methods of finding the inverse system of an LTI system.
Solve:
a)
令 y ( t ) = x ( t ) ∗ h ( t ) (1-1) y(t)=x(t)*h(t)\tag{1-1} y(t)=x(t)∗h(t)(1-1)
若要加一系统 h 1 ( t ) h_{1}(t) h1​(t),使得：
y ( t ) ∗ h 1 ( t ) = x ( t ) (1-2) y(t)*h_{1}(t)=x(t) \tag{1-2} y(t)∗h1​(t)=x(t)(1-2)
则需：
h ( t ) ∗ h 1 ( t ) = δ ( t ) (1-3) h(t)*h_{1}(t)=\delta(t)\tag{1-3} h(t)∗h1​(t)=δ(t)(1-3)
故有：
h 1 ( t ) = ∑ k = 0 + ∞ e k T δ ( t + k T ) (1-4) h_{1}(t)= \sum_{k=0}^{+\infty} e^{kT} \delta(t+ kT) \tag{1-4} h1​(t)=k=0∑+∞​ekTδ(t+kT)(1-4)

b)
同理，若要恢复 x ( t ) x(t) x(t),则应该将 h 2 ( t ) h_{2}(t) h2​(t) 设计为：
h 1 ( t ) = ∑ k = 0 K 1 a k δ ( t + k T ) (1-5) h_{1}(t)= \sum_{k=0}^{K} \frac{1}{a_{k}} \delta(t+ kT) \tag{1-5} h1​(t)=k=0∑K​ak​1​δ(t+kT)(1-5)

但此时 h 2 ( t ) h_{2}(t) h2​(t) 并不是一个因果系统。所以，不可设计一个因果系统恢复 x ( t ) x(t) x(t) 。

c)
在时域上寻找一个系统 h ( t ) h(t) h(t) 的逆系统 h i n v ( t ) h_{inv}(t) hinv​(t) ,则寻找依据是：
h ( t ) ∗ h i n v ( t ) = δ ( t ) (1-6) h(t)*h_{inv}(t)=\delta(t) \tag{1-6} h(t)∗hinv​(t)=δ(t)(1-6)
而在频域上寻找一个系统 h ( t ) h(t) h(t) 的逆系统 h i n v ( t ) h_{inv}(t) hinv​(t) ,寻找依据是：
H ( ω ) H i n v ( ω ) = 1 (1-7) H(\omega)H_{inv}(\omega)=1 \tag{1-7} H(ω)Hinv​(ω)=1(1-7)

3.In communication system, single-banded modulation is an effective way to save the spectrum resource. Figure 1 shows the structure of the system that realizes the single-banded modulation.

(a)Please give an example to verify the system can realize single-banded modulation.

(b)Design a system that can recover f ( t ) f(t) f(t) from y ( t ) y(t) y(t).

Slove:
a) 考虑如下信号：

根据上图关系有：
y 3 ( t ) = c o s ( w 0 t ) f ( t ) (1-1) y_{3}(t)=cos(w_{0}t)f(t) \tag{1-1} y3​(t)=cos(w0​t)f(t)(1-1)
y 1 ( t ) = f ( t ) ∗ h ( t ) (1-2) y_{1}(t)=f(t)*h(t) \tag{1-2} y1​(t)=f(t)∗h(t)(1-2)
y 2 ( t ) = s i n ( w 0 t ) y 1 ( t ) (1-3) y_{2}(t)=sin(w_{0}t)y_{1}(t) \tag{1-3} y2​(t)=sin(w0​t)y1​(t)(1-3)
y ( t ) = y 2 ( t ) + y 3 ( t ) (1-4) y(t)=y_{2}(t)+y_{3}(t) \tag{1-4} y(t)=y2​(t)+y3​(t)(1-4)

求解 Y 3 ( ω ) Y_{3}(\omega) Y3​(ω) :

根据相乘性质：
Y 3 ( j ω ) = 1 2 π ( π δ ( ω − ω 0 ) + π δ ( ω + ω 0 ) ) ∗ F ( ω ) (1-5) Y_{3}(j \omega)=\frac{1}{2\pi}(\pi\delta(\omega-\omega_{0})+\pi\delta(\omega+\omega_{0}))*F(\omega)\tag{1-5} Y3​(jω)=2π1​(πδ(ω−ω0​)+πδ(ω+ω0​))∗F(ω)(1-5)化简后即为：

Y 3 ( j ω ) = 1 2 ( F ( ω − ω 0 ) + F ( ω + ω 0 ) ) (1-6) Y_{3}(j \omega)=\frac{1}{2}(F(\omega-\omega_{0})+F(\omega+\omega_{0})) \tag{1-6} Y3​(jω)=21​(F(ω−ω0​)+F(ω+ω0​))(1-6)
频谱如下：

求解 Y 1 ( j ω ) Y_{1}(j\omega) Y1​(jω):
Y 1 ( j ω ) = H ( j ω ) F ( j ω ) (1-7) Y_{1}(j\omega)=H(j\omega)F(j\omega) \tag{1-7} Y1​(jω)=H(jω)F(jω)(1-7)

考虑如下信号：
s g n ( t ) = 2 u ( t ) − 2 (1-8) sgn(t)=2u(t)-2 \tag{1-8} sgn(t)=2u(t)−2(1-8)
图像如下：

已知：
u ( t ) ↔ F T 1 j ω + π δ ( ω ) (1-9) u(t) \xleftrightarrow[]{FT}\frac{1}{j\omega}+\pi\delta(\omega) \tag{1-9} u(t)FT ​jω1​+πδ(ω)(1-9)

根据FT的线性性质，有：
S ( ω ) = 2 j ω + 2 π δ ( ω ) − 2 π δ ( ω ) = 2 j ω (1-10) S(\omega)=\frac{2}{j\omega}+2\pi\delta(\omega) -2\pi\delta(\omega)=\frac{2}{j\omega}\tag{1-10} S(ω)=jω2​+2πδ(ω)−2πδ(ω)=jω2​(1-10)

根据对偶性：
S ( t ) = 2 j t ↔ F T 2 π s g n ( − ω ) = 2 π s g n ( ω ) (1-11) S(t)=\frac{2}{jt}\xleftrightarrow[]{FT}2\pi sgn(-\omega)=2\pi sgn(\omega) \tag{1-11} S(t)=jt2​FT ​2πsgn(−ω)=2πsgn(ω)(1-11)
化简整理可得：
h ( t ) = 1 π t ↔ F T − j s g n ( w ) = H ( ω ) (1-12) h(t)=\frac{1}{\pi t}\xleftrightarrow[]{FT}-jsgn(w)=H(\omega) \tag{1-12} h(t)=πt1​FT ​−jsgn(w)=H(ω)(1-12)
图像如下：

故 Y 1 ( ω ) Y_{1}(\omega) Y1​(ω) 图像如下：

求解 Y 2 ( ω ) Y_{2}(\omega) Y2​(ω):
Y 2 ( ω ) = 1 2 π Y 1 ( ω ) ∗ π j ( δ ( ω − ω 0 ) − δ ( ω + ω 0 ) ) (1-13) Y_{2}(\omega)=\frac{1}{2\pi} Y_{1}(\omega)*\frac{\pi}{j}(\delta(\omega-\omega_{0})-\delta(\omega+\omega_{0})) \tag{1-13} Y2​(ω)=2π1​Y1​(ω)∗jπ​(δ(ω−ω0​)−δ(ω+ω0​))(1-13)

图像如下：

求解 Y ( ω ) Y(\omega) Y(ω):
Y ( ω ) = Y 2 ( ω ) + Y 3 ( ω ) (1-14) Y(\omega)=Y_{2}(\omega)+Y_{3}(\omega) \tag{1-14} Y(ω)=Y2​(ω)+Y3​(ω)(1-14)

图像如下：

所以，可以看出，对于输入 x ( t ) x(t) x(t) 信号，输出的频谱是输入频谱调制并去除冗余部分的结果。

b) 先与一个频谱为 H ( ω ) = δ ( ω − ω 0 ) + δ ( ω + ω 0 ) H(\omega)=\delta(\omega-\omega_{0})+\delta(\omega+\omega_{0}) H(ω)=δ(ω−ω0​)+δ(ω+ω0​) 的系统在频域卷积，得到频谱 Y 4 ( ω ) Y_{4}(\omega) Y4​(ω),图如下：

将其再经过一个低通滤波器 H 1 ( ω ) H_{1}(\omega) H1​(ω) 即可，其中：

即可恢复出输入信号。
4.The signal in Figure 2a is a band limited signal, which is also called as bandpass signal. Figure 2b is the sampling and reconstruction system, where
h 1 ( t ) = ∑ l = − ∞ + ∞ δ ( t − l T s ) h_{1}(t)= \sum_{l=-\infty}^{+\infty}\delta(t- lT_{s}) h1​(t)=l=−∞∑+∞​δ(t−lTs​)

(a)According to the Nyquist sampling theorem, what condition should the sampling angular frequency satisfy?

(b)If the filter H ( ω ) H(\omega) H(ω) is shown in Figure 2c, ω 1 = ω 2 − ω 1 \omega_{1}=\omega_{2}-\omega_{1} ω1​=ω2​−ω1​ If the sampling angular frequency ω s \omega_{s} ωs​ doesn’t satisfy the Nyquist sampling theorem, can f ( t ) f(t) f(t) be covered from y ( t ) y(t) y(t) ? If yes, please determine ω s \omega_{s} ωs​

Solve:
a)
ω M = ω 2 (1-1) \omega_{M}=\omega_{2}\tag{1-1} ωM​=ω2​(1-1)
ω s ≥ 2 ω M = 2 ω 2 (1-2) \omega_{s} \geq 2\omega_{M}=2\omega_{2} \tag{1-2} ωs​≥2ωM​=2ω2​(1-2)

b)
根据 p ( t ) p(t) p(t) 的表达式，可以得到 P ( ω ) P(\omega) P(ω) 为：

若采样信号不满足奈奎斯特采样定理，即 ω s ≤ 2 ω 2 \omega_{s} \leq 2\omega_{2} ωs​≤2ω2​,
其中 ω 2 = 2 ω 1 \omega_{2}=2\omega_{1} ω2​=2ω1​ ,假设 ω s = ω 2 = 2 ω 1 \omega_{s}=\omega_{2}=2\omega_{1} ωs​=ω2​=2ω1​,则有 Y 1 ( ω ) Y_{1}(\omega) Y1​(ω) 图像如下：

通过上图滤波器后，即得到输入信号，此时 ω s = ω 2 = 2 ω 1 \omega_{s}=\omega_{2}=2\omega_{1} ωs​=ω2​=2ω1​