Hey all,

My current I.F. bandwidth is 8mhz at the 6db points. I am looking at pulses
of .8microseconds length, or about 1.25mhz. If all else remains the same,
and I change the swamping resistors, and tweak the slugs for a 1.5mhz I.F.
bandwidth at the 6db points, what increase in signal to noise ratio should
I see?

TNX 73

Is this an exam question?

If "tweaking" the slugs for a more than six times reduction in bandwidth
doesn't break anything your signal/noise should go up by around 8dB --
you'll lose than 8dB of noise, but you'll also lose a bit of the signal's
sidebands.

Your pulses will be way distorted, though.

Did I pass?

"gudmundur" wrote in message
...
Hey all,

My current I.F. bandwidth is 8mhz at the 6db points. I am looking at
pulses
of .8microseconds length, or about 1.25mhz. If all else remains the same,
and I change the swamping resistors, and tweak the slugs for a 1.5mhz I.F.
bandwidth at the 6db points, what increase in signal to noise ratio should
I see?

TNX 73

Is this an exam question?

If "tweaking" the slugs for a more than six times reduction in bandwidth
doesn't break anything your signal/noise should go up by around 8dB --
you'll lose than 8dB of noise, but you'll also lose a bit of the signal's
sidebands.

Your pulses will be way distorted, though.

Did I pass?

"gudmundur" wrote in message
...
Hey all,

My current I.F. bandwidth is 8mhz at the 6db points. I am looking at
pulses
of .8microseconds length, or about 1.25mhz. If all else remains the same,
and I change the swamping resistors, and tweak the slugs for a 1.5mhz I.F.
bandwidth at the 6db points, what increase in signal to noise ratio should
I see?

TNX 73

In article ,
(gudmundur) writes:

My current I.F. bandwidth is 8mhz at the 6db points. I am looking at pulses
of .8microseconds length, or about 1.25mhz. If all else remains the same,
and I change the swamping resistors, and tweak the slugs for a 1.5mhz I.F.
bandwidth at the 6db points, what increase in signal to noise ratio should
I see?

Signal to noise ratio changes as the _square_root_ of bandwidth
change. Wouldn't be much of an effect going from 1.25 to 1.5 MHz.

With 0.8 uSec pulses and a 1.25 to 1.5 MHz bandwidth (I presume
Mega Hertz, not milli Hertz), the output envelope will be very
rounded, almost Guassian or "cosine-quared" in shape. Rounding
happens because of the limitation of passing the harmonics of the
pulsed RF; all you have left is the carrier frequency.

The relatively narrow bandpass and pulse rounding MAY be okay
in your application. It wasn't stated. I spent some years on a
program that deliberately used 1 MHz bandwidth filters for 1 uSec
wide pulses on carriers 1 MHz apart. Interesting to see the effect
of "matched filters" on adjacent frequencies...those immediately
on each side came through with a reduced amplitude "bow tie"
envelope shape, the "knot" in the middle.

Len Anderson
retired (from regular hours) electronic engineer person

In article ,
(gudmundur) writes:

My current I.F. bandwidth is 8mhz at the 6db points. I am looking at pulses
of .8microseconds length, or about 1.25mhz. If all else remains the same,
and I change the swamping resistors, and tweak the slugs for a 1.5mhz I.F.
bandwidth at the 6db points, what increase in signal to noise ratio should
I see?

Signal to noise ratio changes as the _square_root_ of bandwidth
change. Wouldn't be much of an effect going from 1.25 to 1.5 MHz.

With 0.8 uSec pulses and a 1.25 to 1.5 MHz bandwidth (I presume
Mega Hertz, not milli Hertz), the output envelope will be very
rounded, almost Guassian or "cosine-quared" in shape. Rounding
happens because of the limitation of passing the harmonics of the
pulsed RF; all you have left is the carrier frequency.

The relatively narrow bandpass and pulse rounding MAY be okay
in your application. It wasn't stated. I spent some years on a
program that deliberately used 1 MHz bandwidth filters for 1 uSec
wide pulses on carriers 1 MHz apart. Interesting to see the effect
of "matched filters" on adjacent frequencies...those immediately
on each side came through with a reduced amplitude "bow tie"
envelope shape, the "knot" in the middle.

Len Anderson
retired (from regular hours) electronic engineer person

(Avery Fineman) wrote in message ...
In article ,
(gudmundur) writes:

My current I.F. bandwidth is 8mhz at the 6db points. I am looking at pulses
of .8microseconds length, or about 1.25mhz. If all else remains the same,
and I change the swamping resistors, and tweak the slugs for a 1.5mhz I.F.
bandwidth at the 6db points, what increase in signal to noise ratio should
I see?

Signal to noise ratio changes as the _square_root_ of bandwidth
change. Wouldn't be much of an effect going from 1.25 to 1.5 MHz.

Um, he was starting with an 8MHz BW...

With 0.8 uSec pulses and a 1.25 to 1.5 MHz bandwidth (I presume
Mega Hertz, not milli Hertz), the output envelope will be very
rounded, almost Guassian or "cosine-quared" in shape. Rounding
happens because of the limitation of passing the harmonics of the
pulsed RF; all you have left is the carrier frequency.

Yes, the pulses will certainly be rounded when they come out of the
filter (though they may have started that way anyway). But depending
on the filter type, they may also incur lots of ringing, and if the
pulses follow one after another at the right spacing, the phase of the
energy in the pulse relative to the phase of the energy left in the
filter will matter a whole lot in what you see coming out. The
trailing edge of a rectangular pulse fed through a Chebychev filter
isn't very Gaussian looking!

As for the original question, the answer depends on the spectral
distribution of the noise...if it happens to be strongly peaked at the
carrier frequency of the pulses, the narrowing won't make much
difference; if it happens to be peaked at some other frequency, it may
help a lot. If it's uniformly distributed, AND you keep the filter
shape the same and narrow the bandwidth by 1.5:8, then you will have
1.5/8 as much noise _power_. You'll also have somewhat less signal
power, depending on the shape of the pulses. And of course, the
filter won't get rid of noise that's introduced after the
filter--fairly obvious but sometimes overlooked.

Cheers,
Tom

(Avery Fineman) wrote in message ...
In article ,
(gudmundur) writes:

My current I.F. bandwidth is 8mhz at the 6db points. I am looking at pulses
of .8microseconds length, or about 1.25mhz. If all else remains the same,
and I change the swamping resistors, and tweak the slugs for a 1.5mhz I.F.
bandwidth at the 6db points, what increase in signal to noise ratio should
I see?

Signal to noise ratio changes as the _square_root_ of bandwidth
change. Wouldn't be much of an effect going from 1.25 to 1.5 MHz.

Um, he was starting with an 8MHz BW...

With 0.8 uSec pulses and a 1.25 to 1.5 MHz bandwidth (I presume
Mega Hertz, not milli Hertz), the output envelope will be very
rounded, almost Guassian or "cosine-quared" in shape. Rounding
happens because of the limitation of passing the harmonics of the
pulsed RF; all you have left is the carrier frequency.

Yes, the pulses will certainly be rounded when they come out of the
filter (though they may have started that way anyway). But depending
on the filter type, they may also incur lots of ringing, and if the
pulses follow one after another at the right spacing, the phase of the
energy in the pulse relative to the phase of the energy left in the
filter will matter a whole lot in what you see coming out. The
trailing edge of a rectangular pulse fed through a Chebychev filter
isn't very Gaussian looking!

As for the original question, the answer depends on the spectral
distribution of the noise...if it happens to be strongly peaked at the
carrier frequency of the pulses, the narrowing won't make much
difference; if it happens to be peaked at some other frequency, it may
help a lot. If it's uniformly distributed, AND you keep the filter
shape the same and narrow the bandwidth by 1.5:8, then you will have
1.5/8 as much noise _power_. You'll also have somewhat less signal
power, depending on the shape of the pulses. And of course, the
filter won't get rid of noise that's introduced after the
filter--fairly obvious but sometimes overlooked.

Cheers,
Tom

I went from 8 mhz to 1.5 mhz,,,,
As for breaking things, the wide i.f. was stagger tuned to achieve bandwidth
and swamped with resistors, I could chop the resistors, and retune, and get
25khz bandwidth if I wanted to. It is a short pulse radar i.f., and it will
be used in a long pulse application,,,

Therefore going from 8mhz at 6db edges to 1.5mhz at 6db edges is a must. Any
narrower, and I lose object resolution, any wider, and I amplify unwanted
and detrimental noise.


In article ,
says...

In article ,
(gudmundur) writes:

My current I.F. bandwidth is 8mhz at the 6db points. I am looking at pulses
of .8microseconds length, or about 1.25mhz. If all else remains the same,
and I change the swamping resistors, and tweak the slugs for a 1.5mhz I.F.
bandwidth at the 6db points, what increase in signal to noise ratio should
I see?

Signal to noise ratio changes as the _square_root_ of bandwidth
change. Wouldn't be much of an effect going from 1.25 to 1.5 MHz.

With 0.8 uSec pulses and a 1.25 to 1.5 MHz bandwidth (I presume
Mega Hertz, not milli Hertz), the output envelope will be very
rounded, almost Guassian or "cosine-quared" in shape. Rounding
happens because of the limitation of passing the harmonics of the
pulsed RF; all you have left is the carrier frequency.

The relatively narrow bandpass and pulse rounding MAY be okay
in your application. It wasn't stated. I spent some years on a
program that deliberately used 1 MHz bandwidth filters for 1 uSec
wide pulses on carriers 1 MHz apart. Interesting to see the effect
of "matched filters" on adjacent frequencies...those immediately
on each side came through with a reduced amplitude "bow tie"
envelope shape, the "knot" in the middle.

Len Anderson
retired (from regular hours) electronic engineer person

I went from 8 mhz to 1.5 mhz,,,,
As for breaking things, the wide i.f. was stagger tuned to achieve bandwidth
and swamped with resistors, I could chop the resistors, and retune, and get
25khz bandwidth if I wanted to. It is a short pulse radar i.f., and it will
be used in a long pulse application,,,

Therefore going from 8mhz at 6db edges to 1.5mhz at 6db edges is a must. Any
narrower, and I lose object resolution, any wider, and I amplify unwanted
and detrimental noise.


In article ,
says...

In article ,
(gudmundur) writes:

My current I.F. bandwidth is 8mhz at the 6db points. I am looking at pulses
of .8microseconds length, or about 1.25mhz. If all else remains the same,
and I change the swamping resistors, and tweak the slugs for a 1.5mhz I.F.
bandwidth at the 6db points, what increase in signal to noise ratio should
I see?

Signal to noise ratio changes as the _square_root_ of bandwidth
change. Wouldn't be much of an effect going from 1.25 to 1.5 MHz.

With 0.8 uSec pulses and a 1.25 to 1.5 MHz bandwidth (I presume
Mega Hertz, not milli Hertz), the output envelope will be very
rounded, almost Guassian or "cosine-quared" in shape. Rounding
happens because of the limitation of passing the harmonics of the
pulsed RF; all you have left is the carrier frequency.

The relatively narrow bandpass and pulse rounding MAY be okay
in your application. It wasn't stated. I spent some years on a
program that deliberately used 1 MHz bandwidth filters for 1 uSec
wide pulses on carriers 1 MHz apart. Interesting to see the effect
of "matched filters" on adjacent frequencies...those immediately
on each side came through with a reduced amplitude "bow tie"
envelope shape, the "knot" in the middle.

Len Anderson
retired (from regular hours) electronic engineer person

On 05 Feb 2004 17:35:31 GMT, (Avery Fineman)
wrote:

Signal to noise ratio changes as the _square_root_ of bandwidth
change.

I do not quite understand this.

Usually the signal to noise ratio is defined as the signal power S
compared to the noise N (or compared to S+N). In a white noise
environment with constant noise density, the noise power is directly
proportional to bandwidth.

If the noise bandwidth is larger than the required signal bandwidth,
reducing the noise bandwidth will not affect the signal but only
reduce the noise power directly proportional to the bandwidth
(dropping the bandwidth to 1/4 will drop the noise power by 1/4 or 6
dB and increase the SNR by 6 dB).

However, if you looking at the noise _voltage_, it will drop by the
square_root of the bandwidth. But dropping the bandwidth to 1/4 will
drop the noise voltage to 1/2, which is again 6 dB.

In the OP's case, some of the signal sidebands are also cut, thus also
reducing the signal power. Are you assuming something about the
spectral or phase distribution of these sidebands (e.g. adding
coherent side bands produce the sum of the sideband _voltages_, while
adding noise from two side bands with random phase only produces a sum
of sideband _power_).

Paul OH3LWR