Question

Questions:
a) Write down an expression for the voltage wave approaching the junction from the left Vi(z < 0) in terms of its initial amplitude, Vo, w, k = w/v and z. Here v is the velocity of the wave.

b) What are the voltage amplitude (relative to Vi) of the reflected and transmitted waves for Z(L) = 0.

In what follows (in all 3 questions). Z(L) = 100 ohm.

c) What in the net impedance which terminates the first transmission line at z=0.

d) What are the amplitudes of the reflected and transmitted voltages?

e) If the maximum current in the upper wire for <0 is 10 A, what then is the maximum current in

(i) the lower wire for z<0
(ii) the load
(iii) the upper wire for z>0
(iv) the lower wire for z> 0?​

Answer 1

) \bf{ y(t) = y_0sinωt }y(t)=y

0

sinωt

\bf{ Y0 }Y0

\bf{ u(x) = (0, when x < 0 ) }u(x)=(0,whenx<0)

\bf{ \: \: = U_0, when 0 ≤ x ≤ L }=U

0

,when0≤x≤L

\bf{ \: \: = 0, when x > L }=0,whenx>L

\begin{gathered} \\ \end{gathered}

b)

\begin{gathered} \gamma n = \sqrt{2} (1 \times \frac{ {4}^{1} }{4} ) \gamma (for \: low \: frequencies) \\ \\ \\ v ^{2} n = \frac{ {c}^{2} }{2(1 + \frac{ {4}^{1} }{4}) } \end{gathered}

γn=

2

(1×

4

4

1

)γ(forlowfrequencies)

v

2

n=

2(1+

4

4

1

)

c

2

\begin{gathered} \\ \end{gathered}

c) \large\bf \pink{ ⇝}⇝ The input impedance of a transmission line will be its characteristic impedance if the end terminator equals Zo. So, if Zo = RL then the input impedance to the line will be Zo irrespective of length.

\large\bf \purple{⇝}⇝ If RL does not equal Zo then you get problems with line mismatches and reflections and these vary with operating frequency to cause a significant headache for digital transmission systems.

\large\bf \pink{⇝}⇝ So, the assumption is that the terminating resistor RL equals the characteristic impedance Zo. This means that if RS = Zo then you have a simple potential divider.

\begin{gathered} \\ \end{gathered}

d)

\begin{gathered}a = \frac{v + iz}{2 \sqrt{ |re(z)1 \| } } \\ \\ \\ b = \frac{v - iz}{2 \sqrt{ |re(zo)1| } } \\ s = \frac{a}{b} \end{gathered}

a=

2

∣re(z)1∥

v+iz

b=

2

∣re(zo)1∣

v−iz

s=

b

a

\begin{gathered} \\ \end{gathered}

e) i) \bf{ ε \: is \: 0 \: ahead \: of \: i }εis0aheadofi

\bf{because \: i \: and \: R \: is \: in \: phase \: and \: i.e., \: Z }becauseiandRisinphaseandi.e.,Z

\bf{ and \: ε \: are \: in \: phase \: as \: ε=ε_0 sin \: ω \: from \: diagram. }andεareinphaseasε=ε

0

sinωfromdiagra

Answer 2

a) $\bf{y(t)=y-0sinωt}$=y(t)

$\bf{40}$

$\bf{4(X)=(0,whenX<0)}$

$\bf{ \: \: = u_0,when 0≤X≤}=00$

$\bf{ \: \: = 0, when x > < 3 = 0, when xzL }$

$\begin{gathered} \\ \end{gathered}$

b) $\gamma n= \sqrt{2}(1\times\frac{{4}^{1}}{4} \gamma(for \: low \: frequencies ) \\ \\ \\ v^{2}n=\frac{{c}^{2}{2(1+\frac{4}^{1}}{4})}}$

$\begin{\gathered} \\ \end{gathered}$

c) $\large\bf \pink{⇝}$The input impedance of a transmission line will be its characteristic impedance if the end terminator equals to Zo. So, if Zo=RL then the input impedance to the line will be Zo irrespective of length.

$\large\bf \purple{⇝}$If RL does not equal zo then you get problems with line mismatches and reflections and these vary with operating frequency to cause a significant headache for digital transmission systems.

$\large\bf \pink{⇝}$ So, the assumption is that the terminating resistor RL equal the characteristics implies Zo. This means that if Rs=Zo other you have a simple potential divider.

$\begin{gathered} \\ \end{gathered}$

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