Half pulse shaping circuit help needed

[shaikss]

Hi,

I need to design a circuit which gives positive half cycle of sine wave when logic 1 is given as input and negative cycle of sine wave when logic 0 is input. Please let me know how to design the same using cadence.

 

[GUMY]

Basically You have to design sqare wave to sine wave transformation. Do You have to control amplitude and offset of sine wave?

 

[shaikss]

Basically You have to design sqare wave to sine wave transformation. Do You have to control amplitude and offset of sine wave?

I am not controlling amplitude or offset. I simply need transformation through analog blocks. I want to check the simulation in cadence.

 

[rahdirs]

@shaikss: Do u have sine-wave/square wave generator or do u even need to design it with IC 8038.

Assuming u have square-pulse(0V - 5V) & a sine-wane(-1V - 1V). Can u also mention the frequency of ur square wave & sine wave.I assumed that frequency of o/p is such that comparator using 741 & half-wave rectifier using diode works.

Use two comparators such that,when your square wave is higher than (2.5 V) 1st comparator is switched on (2nd off) & when it is lower than 2.5V 2nd comparator is on(1st off).

When sq.wave.in > 2.5 V,1st comparator is on giving logic '1',multiply this with sine wave using IC 633 & use a half-wave rectifier(diode) to give positive pulses of sine wave.
When sq.wave.in < 2.5V ,2nd comparator is on giving logic '1',multiply this with sine wave & use a half wave rectifier in reverse fashion to give negative pulses of sine wave.

U'll have 2 outputs in ur circuit.U'll get positive pulses on one & negative on other.If u want a single output then just add using an adder circuit to get a single output

 

[shaikss]

Hi,

Do u have sine-wave/square wave generator or do u even need to design it with IC 8038.

Assuming u have square-pulse(0V - 5V) & a sine-wane(-1V - 1V).

Use two comparators such that,when your square wave is higher than (2.5 V) 1st comparator is switched on (2nd off) & when it is lower than 2.5V 2nd comparator is on(1st off).

When sq.wave.in > 2.5 V,1st comparator is on giving logic '1',multiply this with sine wave using IC 633 & use a half-wave rectifier(diode) to give positive pulses of sine wave.
When sq.wave.in < 2.5V ,2nd comparator is on giving logic '1',multiply this with sine wave & use a half wave rectifier in reverse fashion to give negative pulses of sine wave.

U'll have 2 outputs in ur circuit.U'll get positive pulses on one & negative on other.If u want a single output then just add using an adder circuit to get a single output

I have to simulate it using cadence. I have digital data at 1Mbps. So, for simulation purpose, I can make use of pulse data with a period of 1 us. But the problem is how to convert it to sine?

 

[keith1200rs]

I think a Bessel low pass filter will convert a rectangular pulse into a reasonable approximation of a half sine pulse which may help. A sine squared filter also works but is not so easy to make.

Keith

 

[rahdirs]

I have to simulate it using cadence. I have digital data at 1Mbps. So, for simulation purpose, I can make use of pulse data with a period of 1 us. But the problem is how to convert it to sine?

1MHz!!!! and i assumed that the frequency of your sine-wave will not be greater than 1kHz.
Maybe you could do a square to sine-wave transformation.
Convert square-wave to triangle-wave and then convert it to sine the way they do it in 8038.

But i think this method is long & maybe there are better ways to do it.