Signals and Systems Test 2

Concept:

For a given Fourier series coefficient c_n, the signal is given by:

\(x\left( t \right) = \mathop \sum \limits_{n = - \infty }^\infty {c_n}{e^{jn{\omega _0}t}}\)

Expanding the above, we can write:

\(x\left( t \right) = \ldots + {c_{ - 1}}{e^{ - j{\omega _0}t}} + {c_0} + {c_1}{e^{j{\omega _0}t}} + \ldots \) 

ω₀ = fundamental frequency

For ω₀ = 2, the above expansion can be written as:

\(x\left( t \right) = \ldots + {c_{ - 1}}{e^{ - j2t}} + {c_0} + {c_1}{e^{j2t}} + \ldots \)    ---(1)

Application:

\({\sin ^2}t = \frac{1}{2} - \frac{{\cos 2t}}{2}\)

∴ The fundamental period of sin² t will be:

\({T_0} = \frac{{2\pi }}{2} = \pi \) 

And the fundamental frequency ω₀ = 2

sin²t can be expressed as:

\({\sin ^2}t = {\left( {\frac{{{e^{jt}} - {e^{ - jt}}}}{{2j}}} \right)^2}\) 

\({\sin ^2}t = \frac{{ - 1}}{4}\left( {{e^{2jt}} - 2 + {e^{ - 2jt}}} \right)\;\)      ---(2)

Comparing equation (2) with the standard expression of Equation (1), we can write:

\({c_n} = \frac{{ - 1}}{4}\delta \left( {n - 1} \right) + \frac{1}{2}\delta \left( n \right) - \frac{1}{4}\delta \left( {n + 1} \right)\;\;\) 

Time scaling will not effect the Fourier series coefficient

x (0.5t) → C_k

\(x\left( {t - 0.5} \right) \to {e^{ - j{\omega _0}0.5k}}{C_k}\)

x (2t) → C_k

∴ Fourier series coefficient of the given signal is

(-1)ⁿ can be written as (e^jπ)ⁿ

y(n) can now be written as:

\({\left( { - 1} \right)^n}x\left[ n \right] = {\left( {{e^{j\pi }}} \right)^n}x\left[ n \right]\)

This can also be written as:

\(y\left[ n \right] = {e^{j\left( {\frac{{2\pi }}{N}} \right)\left( {\frac{N}{2}} \right)n}}x\left[ n \right]\)

The Fourier series coefficient of y[n] will be:

\({d_k} = \frac{1}{N}\;\mathop \sum \limits_{n = 0}^{N - 1} x\left[ n \right]{e^{j\left( {\frac{{2\pi }}{N}} \right)\frac{N}{2}n\;}}{e^{ - j\left( {\frac{{2\pi }}{N}} \right)nk}}\)

\(= \frac{1}{N}\;\mathop \sum \limits_{n = 0}^{N - 1} x\left[ n \right]{e^{ - j\left( {\frac{{2\pi }}{N}} \right)n\left( {k - \frac{N}{2}} \right)}}\)

\({d_k} = {c_{k - \frac{N}{2}}}\)

x(t) can be written as:

x(t) = x₁(t) + x₂(t)

\({x_1}\left( t \right) = \cos \left( {\frac{{2\pi }}{3}t} \right)\) 

\({x_2}\left( t \right) = 2\cos \left( {\frac{{5\pi }}{3}t} \right)\) 

The time period of x₁(t) is:

\({T_1} = \frac{{2\pi }}{{2\pi /3}} = 3\) 

Time period of x₂(t) is:

\({T_2} = \frac{{2\pi \;}}{{5\pi /3}} = \frac{6}{5}\) 

Time period of x(t) will be:

\({T_0} = LCM\;\left( {{T_1},\;{T_2}} \right) = 6\) 

∴ The fundamental frequency will be:

\({\omega _0} = \frac{{2\pi }}{{{T_0}}} = \frac{{2\pi }}{6} = \frac{\pi }{3}\) 

Now, in exponential terms, x(t) can be written as:

\(x\left( t \right) = \frac{1}{2}{e^{j \cdot \frac{{2\pi }}{3} \cdot t\;}} + \frac{1}{2}{e^{ - j\frac{{2\pi }}{3}t}} + {e^{j\frac{{5\pi }}{3}t}} + {e^{ - j\frac{{5\pi }}{3}t}}\) 

In terms of the fundamental frequency ω₀, we can write:

\(x\left( t \right) = \frac{1}{2}{e^{j2{\omega _0}t}} + \frac{1}{2}{e^{ - j2{\omega _0}t}} + {e^{j5{\omega _0}t}} + {e^{ - j5{\omega _0}t}}\) 

Also, the general expression for the exponential Fourier series is written as:

\(x\left( t \right) = \mathop \sum \limits_{n = - \infty }^\infty {c_n}{e^{jn{\omega _0}t}}\) 

Comparing x(t) with the above general expression we get:

c_-5 = 1

\({c_{ - 2}} = \frac{1}{2}\) 

\({c_2} = \frac{1}{2}\)

c₅ = 1

∴ The amplitude of the second harmonic component will be:

\({c_2} = \frac{1}{2}\) 

Fourier series coefficient is given as:

\({c_k} = \frac{1}{N}\mathop \sum \limits_{\left\langle N \right\rangle } x\left( n \right){e^{ - j\left( {\frac{{2\pi }}{N}} \right)kn}}\)

for k = 0:

\({c_0} = \frac{1}{N}\mathop \sum \limits_{\left\langle N \right\rangle } x\left[ n \right]\)

So, If x(n) is real, c₀ will also be real.

For \(k = \frac{N}{2}\), we can write:

\({c_k} = \frac{1}{N}\mathop \sum \limits_{\left\langle N \right\rangle } x\left( n \right){e^{ - j\left( {\frac{{2\pi }}{N}} \right)\left( {\frac{N}{2}} \right)\;n}}\)

\(\because {e^{ - j\pi n}} = {\left( { - 1} \right)^n}\)

\({c_{N/2}} = \frac{1}{N}\mathop \sum \limits_{\left\langle N \right\rangle } x\left[ n \right]{\left( { - 1} \right)^n}\)

So, If x[n] is real, c_N/2 will also be real.

∴ We conclude that at least two of the FS coefficient are real within one period, If N is even.

Concept:

Properties of Fourier series coefficient:

1) If x(t) is real, then the Fourier series coefficient satisfies the condition:

\({c_n} = c_{ - n}^*\) (* denotes conjugate)

2) If x(t) is even, then:

c_n = c_-n

3) Fourier series coefficient of \(\frac{{dx\left( t \right)}}{{dt}}\) is given by:

\(\frac{{dx\left( t \right)}}{{dt}}\mathop \leftrightarrow \limits^{C.T.F.S} jn\frac{{2\pi }}{{{T_0}}} \cdot {c_n}\)

Application:

Given:

\({c_n} = \left\{ {\begin{array}{*{20}{c}} {2,\;\;\;\;\;\;\;,\;\;\;\;\;\;\;\;\;~n = 0}\\ {j{{\left( {\frac{1}{2}} \right)}^{\left| n \right|}},\;\;\;\;otherwise} \end{array}} \right.\)

c_-n will be defined as:

\(c_{ - n}^* = \left\{ {\begin{array}{*{20}{c}} {2\;\;\;\;\;\;\;\;\;,\;\;\;\;\;\;n = 0}\\ { - j{{\left( {\frac{1}{2}} \right)}^{\left| n \right|}},\;\;\;\;otherwise} \end{array}} \right.\)

Since \({c_n} \ne c_{ - n}^*\), x(t) is not real.

Also, we observe that c_n = c_-n.

∴ The function x(t) is even.

\(x\left( t \right) \leftrightarrow {c_n}\)

\(\frac{{dx\left( t \right)}}{{dt}}\mathop \leftrightarrow \limits^{C.T.F.S} c_n' = jn\frac{{2\pi }}{{{T_0}}} \cdot {c_n}\)

\(c_n' = \left\{ {\begin{array}{*{20}{c}} {0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;,\;\;\;\;\;\;n = 0}\\ { - n{{\left( {\frac{1}{2}} \right)}^{\left| n \right|}}.\frac{{2\pi }}{{{T_0}}}\;\;,\;\;\;\;otherwise} \end{array}} \right.\)

We observe that \(c_n' \ne c_{ - n}'\)

\(\therefore y\left( t \right) = \frac{{dx\left( t \right)}}{{dt}}\) is not even.