This post consists of solutions to problems from 1.1 to 1.21 in chapter 1 of the book Signals and Systems by Oppenheim ,Willsky and Nawab.
Problem 1.1: Express each of the following complex numbers in Cartesian form \(x+iy\):
First, let’s figure them out using the Euler’s formula. For the nine complex numbers in Problem 1.1, we have:
\begin \tfrac e^ &=& \tfrac (\cos(\pi) + i\sin(\pi) ) = \tfrac \cos(\pi) = -\tfrac \\
\tfrac e^ <-i\pi>&=& \tfrac (\cos(-\pi) + i\sin(-\pi) ) = \tfrac \cos(-\pi) = -\tfrac \\
e^ &=& \cos(\pi/2) + i \sin(\pi/2) = i \\
e^ <-i\pi/2>&=& \cos(-\pi/2) + i \sin(-\pi/2) = -i \\
e^
\sqrte^ &=& \sqrt ( \cos(\pi/4) + i \sin(\pi/4) ) = \sqrt ( \tfrac<\sqrt> + i\tfrac<\sqrt> ) = 1 + i \\
\sqrt e^ &=& \sqrt ( \cos(9\pi/4) + i \sin(9\pi/4) ) = 1 + i \\
\sqrt e^ &=& \sqrt ( \cos(-9\pi/4) + i \sin(-9\pi/4) ) = 1 - i \\
\sqrt e^ &=& \sqrt ( \cos(-\pi/4) + i \sin(-\pi/4) ) = 1 - i \end
Problem 1.2: Express each of the following complex numbers in polar form (\(re^, -\pi < \theta \leq \pi\)):
For the nine complex numbers in Problem 1.2 , we have:
Every complex number \(a+ib\) can be visualized in the complex plane \(\mathbb\). It can be viewed either as the point with the coordinate \((a,b)\) or as a vector starting from \(( 0,0 )\) to the point \((a,b)\).
Also every complex number \(a+ib\) can be represented in the exponential form conveniently by using the Euler’s formula \(e^ = \cos\alpha + i\sin\alpha\). One complex number have unique Cartesian form but many expoential forms. Taking \(1+i\) for example, its expoential form can be \(\sqrte^, n\in \< \ldots, -2,-1,0,1,2,\ldots \>\). When we express a complex number in expoential form, it helps to keep a concept of rotation in mind. In the complex plane \(\mathbb\), a complex number will return to itself if it rotates a multiple of \(2\pi\) around the origin point with radius equal to its modulus. please keep the concept of roatation in mind and it will become increasingly important during our later study.
In the development of complex analysis, \(a+ib\) has another form \(r\angle \theta\), where \(r\) is the modulus and \(\theta\) the argument (or the angle). Obviously, this notation is not as good as the expoential form, espacialy when we want to do complex analysis such as differentiation and integration. We just mention it here for the sake of completion. The expoential form will be deployed from now on.
Now, let’s go back to Problem 1.1 and Problem Problem 1.2. It’s easy to figure the answers out using the Euler’s formula. Let’s do more to show them on the complex plane keeping the concept of rotation in mind.
From Figure 1 , taking \(\colore^>\) and \(\colore^<-i\pi>>\) for example, in the complex plane, they are the same point \(-\tfrac\) which means \(-\tfrac\) can be reached by rotating \(\tfrac\) with angle \(\pi\) anti-clockwise or with angle \(-\pi\) clockwise. Essentially, this is because that \(e^ = e^, n\in \<\ldots,-2,-1,0,1,2,\ldots\>\). It’s straightforward that \(e^ = e^ = e^<-i\pi>\).
In the end of Problem 1.1 and Problem 1.2, I want to say more about expressing \(\tfrac\) in its polar form. There are two methods to get the polar form:
Problem 1.3: Determine the values of \(P_<\infty>\) and \(E_<\infty>\) for each of the following signals:
Before solving this problem, let’s review the definition of \(P_<\infty>\) and \(E_<\infty>\). For a continuous time signal \(x(t)\), we have
For a discrete time signal \(x[n]\), we have:
Equation (\ref)(\ref) are not only mathmatical definitions but also related to physical quantities such as power and energy in a physical system. For an electric circuit, taking the voltage \(v(t)\) and current \(i(t)\) across a resistor for example, the power at time \(t\) can be calculated by:
Next let’s determine the values of \(P_<\infty>\) and \(E_<\infty>\) for the given signals.
So, it’s straightforward that:
We have \(|e^| = 1\) , therefore
For power \(P_<\infty>\), we have:
This signal has constant power. If you keep the concept of rotation mentioned in Problem 1, you will notice immediately that all the points generated by \(x_(t)\) lies on the unit circle.
\(x_(t)\) can be visualized as follows:
By definition, we have:
\(x_[n]\) can visualized as:
Problem 1.4: Let \(x[n]\) be a signal with \(x[n]=0\) for\(n< -2\) and \(n>4\), for each signal given below, determine the values of \(n\) for which it is guaranteed to be zero.
For the given signals \(\mathbf\) to \(\mathbf\), the transformations of the variable \(n\) will change the interval in which the signals are zero. For the convenience of calculation, we write the origin signal as:
We can visualize \(x(m)\) as below (I just give an example, you can name any signal that satisfy \(x[m] = 0, m < -2 \ \mathrm\ m>4 \)):
For signal \(\mathbf\), to get the interval where \(x[n-3] = 0 \), we have:
Then, we have \( n < 1 \ \mathrm\ n > 7\) from which we can see that the new signal is a right shift three relative to the origin signal, i.e. new signal is delayed with three.
For signal \(\mathbf\), we have:
Then, we have \(n\ n>0\) from which we can see that the new signal is a left shift four relative to the origin signal, i.e. new signal is advanced with four.
For signal \(\mathbf<( c )>\), we have:
for signal \(\mathbf\), we have:
For signal \(\mathbf\), we have:
\begin \begin \mathbf: x(1-t) \qquad & \mathbf: x(1-t) + x(2-t)\qquad \\
\mathbf<( c)>: x(1-t)x(2-t)\qquad & \mathbf: x(3t) \qquad \\
\mathbf: x(t/3)\qquad & \\
\end \end
For signal \(\mathbf: x(1-t) \), we have:
So, \(t>-2\). Signal \(\mathbf: x(1-t) \) can be achieved by flipping the origin signal first then right shifting one.
For signal \(\mathbf: x(1-t) + x(2-t) \) , we have:
Because it a “\(+\)” between \(x(1-t)\) and \(x(2-t)\) so only when both of the two signals are zero, the result will be zero. For the first part we have the value zero when \(t>-2\) the second \(t>-1\), so the result will be the intersect of these two intervals, i.e. \(t>-1\) . we achieve signal \(\mathbf: x(1-t) + x(2-t) \) by adding signal \(\mathbf: x(1-t) \) with a second signal \(x(2-t)\) which can be achieved by flipping it first then right shifting two.
For signal \(\mathbf<( c)>: x(1-t)x(2-t) \), \(x(1-t)\) is multiplied by \(x(2-t)\), so if any one of these two signals is zero, the result is zero. Based on the result of signal \(\mathbf<( b )>\), we have \(t>-2\).
For signal \(\mathbf<( d )>: x(3t)\), we have:
For signal \(\mathbf<( e )>: x(t/3)\), we have:
Determine whether or not each of the following signals is periodic:
For each signal given below, determine all the values of the independent variable at which the even part of the signal is guaranteed to be zero.
\begin \small \begin \mathbf: x_[n] = u[n] - u[n-4] \qquad & \mathbf: x_(t) = \sin(\tfrac t) \qquad \\
\mathbf<( c)>: x_[n] = (\frac)^ u[n-3] \qquad & \mathbf: x_(t) = e^u(t+2) \\
\end \end
Any signal \(x_(t)\) consists of two parts, even part \(\mathrm \
For signal \(x_(t) = \sin (\frac t)\), because it is an odd signal, so for all \(t\), the even part of \(x_(t)\) is zero.
Notice that for \(\sin(\fract)\), the signal can be obtained by streching \(\sin(t)\) with a factor \(2\). In particular, the fundamental period is \(4\pi\).
Express the real part of each of the following signals in the form \(Ae^\cos(\omega t + \phi)\), where \(A,a,\omega, \phi\) are real numbers with \(A>0\) and \(-\pi < \phi \leq \pi \):
Determine whether or not each of the following signals is periodic. If a signal is periodic, specify its fundamental period.
During figuring all the problems, we have to keep in mind that for a complex number after rotating \(2\pi \) around the origin the number returns to itself.
Determine the fundamental period of the signal \(x(t) = 2\cos(10t + 1) - \sin(4t -1)\)
\(x(t)\) consists of two parts, \(2\cos(10t + 1)\) and \(-\sin(4t -1)\). First, We figure out the fundamental period for these two signals respectively. For \(cos(10t + 1)\), \(10T_ = 2\pi \to T_ = \frac<\pi>\); For \(-\sin(4t -1)\) , we have \( 4T_ = 2\pi \to T_ = \frac<\pi> \) . Then, the fundamental period \(T = mT_ = nT_\) ,So when \(m=5\) and \(n=2\) , we have \(T=\pi\) , \(T\) is the least common multiples of \(T_\) and \(T_\)
Determine the fundamental period of the signal \(x[n] = 1 + e^
\(x[n]\) consists of two main parts, \( e^
For the first part we have:
So when \(m=2\), we have \(N_ = 7\).
For the second part we have:
So when \(m=1\), we have \(N_ = 5\). The lease common multiple of \(N_\) and \(N_\) is 35, i.e. the fundamental period of \(x[n]\)
It is easy to validate it. We have \(x[0] = 1 = x[35]\)
Consider the discrete-time signal
Determine the values of the integers \(M\) and \(n_\) so that \(x[n]\) may be expressed as
By definition of \(\delta[n]\) , we know that when \(n=1+k\), \(\delta[n-1-k] = 1\). For \(x[n]\), because \(k\geq 3\), so for every \(n\geq 4\), there will always be one \(k\) such that \(x[n] = 0\). However, for every \(n\leq 3\), such that \(\sum_^ <\infty>\delta[n-1-k] = 1 \), i.e. \(\sum_^ <\infty>\delta[n-1-k] = 0 \) for all \(n\leq 3\) , So \(x[n] = 1\) for all \(n\leq 3\).
Then, by the definition of unit step function, we have \(M=-1\) and \(n_= -3\) . We can see that \(x[n]\) can be obtained by flipping the unit step signal then right shifting it with \(3\).
Consider the continuous-time signal
\begin x(t) = \delta(t+2) - \delta(t-2) \end
Calculate the value of \(E_<\infty>\) for the signal
First let’s visualize \(x(t)\) which can be shown as below
Notice the “\(1\)” along with the end of the impulses at \((-2,0)\) and \((2,0)\) only represent that the area of the impulse is \(1\).
Consider a periodic signal
with period \(T=2\), The derivative of this signal is related to the “impulse train”
with period \(T=2\). It can be shown that
Determine the values of \(A_,t_,A_\) and \(t_\)
First let’s visualize \(x(t)\) and \(g(t)\) :
Then we try to figure out \(\frac<\mathrm
Compared the result with \(g(t)\), then we have \(A_ = 3\), \(A_ = -3\), \(t_ = 0\) ,\(t_ = 1\)
Consider a system \(S\) with input \(x[n]\) and output \(y[n]\). This system is obtained through a series interconnection of a system \(S_\) followed by a system \(S_\). The input-output relationships for \(S_\) and \(S_\) are:
where \(x_[n]\) and \(x_[n]\) denote input signals.
Because \(S_\) is followed by \(S_\), then the output of \(S_\) is the input of \(S_\) i.e., \(y_[n]\) can be treated as \(x_[n]\) , So,
\(y_\) is the final output of system \(S\), and \(x_[n]\) the input.
If \(S_\) is followed by \(S_\), then \(y_[n]\) can be treated by \(x_[n]\), \(y_[n]\) is the final output of system \(S\) and \(x_[n]\) the input.
After comparing the result from (1) we can see that the two systems are identical, no matter who ( either \(S_\) or \(S_\)) comes first.
Later we will learn that for any number of linear systems, the input-output relationship does not change no matter what order by which they are concatenated.
Consider a discrete-time system with input \(x[n]\) and output \(y[n]\). The input-output relationship for this system is
Consider a continuout-time system with input \(x(t)\) and output \(y(t)\) related by \(y(t) = x( \sin(t) )\).
A system is causal if the output at any time depends on values of the input at only the present and past times. We can anticipate the output of one causal system based on the history of the input. When it comes to \(y(t) = x(\sin(t))\) , we can have \(y(-\pi) = x(0), t=-\pi\). i.e. \(y(-\pi)\) is determined by an input from the future. So the system is not causual.
If a system is linear, we have:
When the input is \(a_x_(\sin(t)) + a_x_(\sin(t)) \), the output will be \(a_y_(t) + a_y_(t)\) . The system is linear.
Consider a discrete-time system with input \(x[n]\) and output \(y[n]\) related by
where \(n_\) is a finite positive integer.
According to definition of linearity, suppose we have two inputs \(x_[n]\) and \(x_[n]\), the reponse is \(y_[n]\) and \(y_[n]\). i.e.
Then for \(x_[n] = x_[n] + x_[n]\), we have
So that the system is linear.
If a system is time invariant, the output does not change if you input the same signal at any time. No matter when you repeat the same experiment, you will get the same output.
For an input \(x_[n]\), we have
Suppose we have another input \(x_[n] = x_[n-n_]\), so
So when we have \(x_[n] = x_[n- n_]\), then we have \(y_[n] = y_[n- n_]\). So, the system is time invariant.
we know that \(x[n] \leq B\), then
So we have \(C \leq ( 2n_ +1 )B\).
For each of the following input-output relationships, determine whether the corresponding system is linear, time invariant or both
Suppose that \(y_(t)\) and \(y_(t)\) are the responses for input \(x_(t)\) and \(x_(t)\), respectively. Given \(x_(t) = ax_(t) + bx_(t)\), then we have:
\begin y_(t) = t^x_(t-1) = t^ \Big( ax_(t-1) + bx_(t-1) \Big) = ay_(t) + by_(t) \end
So the system is linear.
To check whether this system is time invariant, we must determine whether the time-invariant property holds for any input and any time shift \(t_\). Suppose when \(x_(t)\) is applied we have output \(y_(t) = t^x_(t-1)\). We shfit \(x_(t)\) with \(t_\) to get \(x_(t-t_)\) and apply it to the system. Then we have \(y_(t) = t^x_(t-t_-1)\) which is not the shifted version of \(y_(t)\) which is \(y_(t-t_) = (t-t_)^ x_(t-t_-1) \). So the system is not time invariant.
To check whether the system is linear or not, suppose \(x_[n] = ax_[n] + bx_[n]\) then
So the system is not linear
\(x_[n]\) will generate output \(y_[n] = x_^[n-2]\). Suppose \(x_[n] = x_[n-n_]\), then we have \(y_[n] = x_^[n-2] = x_^[n-n_-2] = y_[n-n_]\) , i.e. when the input shift with \(n_\) the output will all shift with \(n_\). So this system is time invariant
First, we check whether or not the system is linear. consider
Let \(x_[n] = ax_[n] + bx_[n]\), then
The system is linear.
Second, we check whether or not the system is time invariant. let \(x_[n] = x_[n-n_]\) then we have
So the system’s output and input see the same time shift. The system is time invariant
First, we check whether or not the system is linear. Considering two inputs \(x_(t)\) and \(x_(t)\), we have:
Let \(x_(t) = ax_(t) + bx_(t)\), then
The system is linear.
Second we check whether or not the system is time-invariant. A system is time invariant if a time shift in the input signal results in an identical shift in the output signal.
If we shift the input \(x(t)\) and get \(x(t-t_)\) then the result \(y(t)\) we become as \(y(t-t_)\). Before we check the property of time-invarant, let’s first check one example. Suppose
We can visualize \(x_(t)\) and \(y_(t)\) as follows:
If we shift \(x_(t)\) by \(2\) to obtain \(x_(t) = x_(t-2)\), then we have
Then we have \(y_(t)\),
Then, we visualize \( y_(t) \),
Notice, by shifting \(y_(t)\) by \(2\) , we have \(y_(t-2)\)
Now, we can derive that \(y_(t-2) \neq y_(t)\) . Hence, The system is not time invariant .
Also we can visualize \(y_(t-2)\) to see how different \(y_(t-2)\) and \(y_(t)\) are.
Now, we deduce that the system is time-invariant. For \(x_(t)\), we have,
Now, let \(x_(t) = x_(t-t_)\) , then
Because \(y_(t) \neq y_(t- t_)\), therefore the system is not time invariant.
A continuous-time linear system \(S\) with input \(x(t)\) and output \(y(t)\) yields the following input-output pairs:
The system \(S\) is linear system, so the output should be
Using Euler’s formula, we represent \(x_(t)\) using the exponential form.
Because \(\frace^\) and \(\frace^\) are complex constants and the system is linear, so we have the output
Therefore we have \(y_(t) = \cos(3(t-\tfrac))\).
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