I think the origin for noise would most often be the channel; however the biggest effect of noise would occur at the receiver after even the noise is amplified.
At the early stage, you have mentioned that the 2 carriers must be out of phase and at the end of the video you say they are in phase!! so which is right?
So the multiplexed signal I am assuming are the data lines? But then you need an input for carrier signal as well right? I assume it would go to the modulator as well?
When you multiply the multiplexed signal s(t) by the carrier wave cos(wc*t) in the receiver's product modulator, you'll see that with the product trig identity the product will be 1/2[Ax(t)cos(2*wc*t) + Ax(t)cos(0)]. Since cos(0) = 1, the second term is just 1/2Ax(t). When you pass that sum through the LPF (low pass filter), the cos(2*wc*t) is attenuated and so the result for the In Phase channel at the top is just [0.5Ax(t)]. This also requires that the LPF has a cutoff frequency greater than the frequency of the original message signal x(t) so that it survives the filter.
@@salukaboy9891 brother since s(t) = x1(t)Acos(wc t) + x2(t)Asin(wc t) and on multiplying wiht carrier wave cos(wc t) we get x1(t)Acos^2(wc t) + x2(t)Asin(wc t)cos(wc t), from this how can we obtain the equation 1/2[Ax(t)cos(2*wc*t) + Ax(t)cos(0)] as you mentioned above?
I think the origin for noise would most often be the channel; however the biggest effect of noise would occur at the receiver after even the noise is amplified.
As shown in Friis formula, the noise figure is primarily set by the noise figure of the transmitter
At the early stage, you have mentioned that the 2 carriers must be out of phase and at the end of the video you say they are in phase!! so which is right?
Good and simple explanation.
Thank you so much...stay connected
Thank you so much sir
Easy explained thanks
thank u so much , j'ai bien compris
You are welcome
So the multiplexed signal I am assuming are the data lines? But then you need an input for carrier signal as well right? I assume it would go to the modulator as well?
Super sir..
Thank you so much ❤😊
Thankyou for this🙌
Sir QAM and QASK are same term na?
Ans: Channel
Sir, QAM modulator transmitter capable of transmitting PAM signals ?
At the transmitter
No, in the channel
I can not understand receiver part, how to {1/2 An+t)} output value.
When you multiply the multiplexed signal s(t) by the carrier wave cos(wc*t) in the receiver's product modulator, you'll see that with the product trig identity the product will be 1/2[Ax(t)cos(2*wc*t) + Ax(t)cos(0)]. Since cos(0) = 1, the second term is just 1/2Ax(t). When you pass that sum through the LPF (low pass filter), the cos(2*wc*t) is attenuated and so the result for the In Phase channel at the top is just [0.5Ax(t)]. This also requires that the LPF has a cutoff frequency greater than the frequency of the original message signal x(t) so that it survives the filter.
@@salukaboy9891 brother since s(t) = x1(t)Acos(wc t) + x2(t)Asin(wc t) and on multiplying wiht carrier wave cos(wc t) we get x1(t)Acos^2(wc t) + x2(t)Asin(wc t)cos(wc t), from this how can we obtain the equation 1/2[Ax(t)cos(2*wc*t) + Ax(t)cos(0)] as you mentioned above?
Channel most likely
at the receiver obviously
No, in the channel friend
In the channel sir...
Yes, correct
QAM reciver diagram is coorect??
Obviously sir in the channel or medium.
Yes
The answer is option D In the channel
Yes...correct
White gossian noise is added in the channel
at the channel mostly sir
Yes..correct dear
Noise is most likely at the transmitter
Yeah Channel has more Chance of Effecting the Noise! We eliminate Noise With Different Modulation Techniques such as AM, FM, PM
Advantage and disadvantages of qam
Ok, I will try to make a video on it.
is x(t) binary or a continuous signal?
X(t) is analog input (continuous signal)
At the cnannel instead please
Yess
noise most likely to occur in channel.
Yes...correct
after the transmission at the channel when the transmitter sends the signal through the medium(channel) to the receiver
Sir , option (in the channel)
through the channel
Right
At the receiver
No friend, in the channel
My answer is D.In the Channel.
Geometric representation of QAM plz.
OK, I will try to make video on it.
D SIR IN THE CHANNEL
Correct
channel is prone to add most of the noise to the signal !!
Yes...correct
In transmission line
D ..in the channel
Correct
Channel only.....
At Receiver
No, in the channel
option D in the channel
please explain how we get 1/2Ax1(t) at the output of qam receiver
transmission channel
Yes
Channel only
in the channel
Correct
D is the answer channel
Right
90degree rotation
Answer is option D
In the Channel
Correct
CHannel
Right👍
d i n the channel
Correct
In the channel
Great video!
Thank you😊
Channel Sir.
D
ఛానెల్
Information source
channel in
Correct
@@EngineeringMadeEasy thank you sir . Your vedios help me a lot and iam very thankful to you .
Sir answer pls
Channel
Monika Chouhan channel is the ryt answer
Channel
C- in the information source
D.channel
ANS : CHANNEL
Correct
Channel..
In a channel
Correct
Comments badhane ki ninja technique 😂
channel
Right answer
Channel
Correct
In channel
Harris Barbara Martin Ronald Hall Donald
OPTION C
You are just reading things from a paper, not explaining.
Channel bro
D channel
B
In the channel
Correct answer
At the receiver
In channel
Right
channel
Yes
Channel
Correct
D
In the channel
Correct
D
Channel
Correct
channel
Yes
In the channel
Correct
Channel
channel
D
In the channel
Right
Channel
channel
Correct
D
In the channel
Right
channel
Right
Channel
Right
D
in the channel
Correct
Channel
Right
channel
In the channel
Correct