A Very lossy $\lambda/4$ long $50\;\Omega$ transmission line is open circuited at the load end. The input impedance measured at the other end of the line is approximately.
a) $0$
b) $50\;\Omega$
c) $\infty$
d) None of these
Correct Answer: Option B
Solution: As soon as we saw $Z_{in}$ for $\lambda/4$ transmission line, we tempt to use the formula, $$Z_{in}=Z_0^2/Z_L$$ $$Z_{in}=50^2/\infty$$ $$Z_{in}=0\;\Omega$$ Hence, we choose the option A.
But the above formula is for lossless$\lambda/4$ transmission line.
In our question, they have asked for very lossy transmission line.
How to approach this question?
Case 1: Lossless transmission line matched with $Z_0$
Since the transmission line is matched, there will be no standing wave and hence $Z_{in}$ will be equal to $Z_0$.
Case 2: Lossless transmission line not matched with $Z_0$
Since the transmission line is not matched, there will be reflections at the load and hence standing wave exists. Whenever standing wave is there, the impedance of the transmission line will vary in accordance voltage and current at that point.
Case 3: Very lossy transmission line not matched with $Z_0$
Since the transmission line is not matched, there will be reflections at the load and hence standing wave exists. But here, standing wave gets attenuated since the transmission line is very lossy. There will be no standing wave at input side. Hence input impedance is equal to characteristic impedance $Z_0$.
A Very lossy $\lambda/4$ long $50\;\Omega$ transmission line is open circuited at the load end. The input impedance measured at the other end of the line is approximately.
a) $0$
b) $50\;\Omega$
c) $\infty$
d) None of these
Correct Answer: Option B
Solution: As soon as we saw $Z_{in}$ for $\lambda/4$ transmission line, we tempt to use the formula, $$Z_{in}=Z_0^2/Z_L$$ $$Z_{in}=50^2/\infty$$ $$Z_{in}=0\;\Omega$$ Hence, we choose the option A.
But the above formula is for lossless$\lambda/4$ transmission line.
In our question, they have asked for very lossy transmission line.
How to approach this question?
Case 1: Lossless transmission line matched with $Z_0$
Since the transmission line is matched, there will be no standing wave and hence $Z_{in}$ will be equal to $Z_0$.
Case 2: Lossless transmission line not matched with $Z_0$
Since the transmission line is not matched, there will be reflections at the load and hence standing wave exists. Whenever standing wave is there, the impedance of the transmission line will vary in accordance voltage and current at that point.
Case 3: Very lossy transmission line not matched with $Z_0$
Since the transmission line is not matched, there will be reflections at the load and hence standing wave exists. But here, standing wave gets attenuated since the transmission line is very lossy. There will be no standing wave at input side. Hence input impedance is equal to characteristic impedance $Z_0$.
The characteristic equation of JK flip-flop is $$Q^+=J\bar{Q}+\bar{K}Q$$ Subs the given inputs, $J=\bar{Q}$ and $K=1$ $$Q^+=\bar{Q}\bar{Q}+0Q$$ $$Q^+=\bar{Q}+0$$ $$Q^+=\bar{Q}$$ From this we can say, for every clock pulse, the output of the JK flip-flop gets toggled. Hence $010101$ is the answer
In a JK flipflop, $J=\bar Q$ and $K=1$. Assuming the flip flop was initially cleared and clocked for $6$ pulses, the sequence at the $Q$ will be
a) $010000$ b) $011001$ c) $010010$ d) $010101$
Correct Answer: Option D
The characteristic equation of JK flip-flop is $$Q^+=J\bar{Q}+\bar{K}Q$$ Subs the given inputs, $J=\bar{Q}$ and $K=1$ $$Q^+=\bar{Q}\bar{Q}+0Q$$ $$Q^+=\bar{Q}+0$$ $$Q^+=\bar{Q}$$ From this we can say, for every clock pulse, the output of the JK flip-flop gets toggled. Hence $010101$ is the answer
CDMA is newest and latest used access method using satellites and popular in _____ applications and not much in ______ applications
a) Commercial, military
b) military, commercial
c) military, defence
d) research, commercial
Correct Answer: Option B
From the book, "Data Communications and Networks" authored by Achyut S Godbole,
CDMA is newest and latest used access method using satellites. It is popular in military installations but not so much in the case of commercial applications.
CDMA is newest and latest used access method using satellites and popular in _____ applications and not much in ______ applications
a) Commercial, military
b) military, commercial
c) military, defence
d) research, commercial
Correct Answer: Option B
From the book, "Data Communications and Networks" authored by Achyut S Godbole,
CDMA is newest and latest used access method using satellites. It is popular in military installations but not so much in the case of commercial applications.
The order in which the following satellite communication is accessed is
a) FDMA TDMA CDMA
b) CDMA FDMA TDMA
c) FDMA CDMA TDMA
d) Cellular Telephony AMPS FDMA
Correct Answer: Option A
We can answer this question using, uplink power requirements
FDMA:
Using,
$$\frac{C}{N_0}=\frac{E_b}{N_0}+R$$
We can say Power level of carrier (C) depends on modulation rate.
In FDMA, we can say modulation rate (R) will be less, as total frequency spectrum is divided among earth stations and hence the EIRP (Equivalent isotropic radiated power) is less for uplink.
TDMA:
We can say TDMA uses modulation rate higher than FDMA and hence higher EIRP is required for the earth stations to uplink.
Hence FDMA is most preferable than TDMA.
Likewise CDMA also has high modulation rate because of generation of pseudo noise like sequence.
The order in which the following satellite communication is accessed is
a) FDMA TDMA CDMA
b) CDMA FDMA TDMA
c) FDMA CDMA TDMA
d) Cellular Telephony AMPS FDMA
Correct Answer: Option A
We can answer this question using, uplink power requirements
FDMA:
Using,
$$\frac{C}{N_0}=\frac{E_b}{N_0}+R$$
We can say Power level of carrier (C) depends on modulation rate.
In FDMA, we can say modulation rate (R) will be less, as total frequency spectrum is divided among earth stations and hence the EIRP (Equivalent isotropic radiated power) is less for uplink.
TDMA:
We can say TDMA uses modulation rate higher than FDMA and hence higher EIRP is required for the earth stations to uplink.
Hence FDMA is most preferable than TDMA.
Likewise CDMA also has high modulation rate because of generation of pseudo noise like sequence.
If calc is not allowed, round off $1001$ to nearest $2^x$ value. Hence $$C\lt 3.4×10^3\; log_2(1024)$$ $$C\lt 3.4×10^3\; log_2(2^{10})$$ $$C\lt 34000\;bps$$
If calc is not allowed, round off $1001$ to nearest $2^x$ value. Hence $$C\lt 3.4×10^3\; log_2(1024)$$ $$C\lt 3.4×10^3\; log_2(2^{10})$$ $$C\lt 34000\;bps$$
